MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 10
Liberty University
Question 1
Question
Samantha invests $5000 in a savings account that earns 4% annual interest
compounded continuously. How much will she have in the account after 10
years?
Solution
Step 1: The formula for compound interest with continuous compounding is
given by:
A=P ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= annual interest rate (in decimal form), t=
time in years, and e= Euler’s number, approximately 2.71828.
Step 2: Given that P= $5000,r= 0.04, and t= 10, we can plug these
values into the formula to find A.
A= 5000 ·e0.04·10
Step 3: Calculate the amount using a calculator:
A≈5000 ·e0.4≈5000 ·1.4918247 ≈7459.12
Step 4: Samantha will have approximately $7459.12 in the account after 10
years.
Question 2
Question
Solve for x:23x+1 −5·2x+1 + 6 = 0.
Solution
Step 1: Let y= 2x. Then, rewrite the equation using y.
23x+1 −5·2x+1 + 6 = 0
⇒2·23x−5·2·2x+ 6 = 0
⇒2·(2x)3−5·2·2x+ 6 = 0
⇒2y3−10y+ 6 = 0
Step 2: Solve the cubic equation by factoring or using any appropriate
method.
Since the coefficient of y3is 2we can simplify the equation by dividing by
2:
y3−5y+ 3 = 0
Step 3: Try potential roots by using the Rational Root Theorem. The
possible rational roots are ±1,±3.
Substitute these values to find the root. We find y= 1 is a root by trial and
error.
Step 4: Use synthetic division to divide y3−5y+ 3 by (y−1).
1 1 0 −5
3
1 1
−4
Step 5: The synthetic division shows that y3−5y+ 3 = (y−1)(y2+y−3).
Step 6: Factor the quadratic polynomial y2+y−3.
y2+y−3 = (y+ 2)(y−1)
Step 7: Set each factor equal to zero and solve for y.
y+ 2 = 0 ⇒y=−2
y−1 = 0 ⇒y= 1
Step 8: Substitute back y= 2xto find the corresponding values of x:
y= 1 ⇒2x= 1 ⇒x= 0
y=−2⇒2x=−2(No real solution)
Step 9: Therefore, the solution to the equation 23x+1 −5·2x+1 + 6 = 0 is
x= 0.
2
Question 3
Question
Samantha invested $10,000 in a savings account that earns 4
Solution
Step 1: The formula for continuously compounded interest is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (initial investment). - ris the annual interest rate
(in decimal form). - tis the time the money is invested for in years. - eis Euler’s
number, approximately equal to 2.71828.
Step 2: Since Samantha wants to double her investment, the amount after t
years will be $20,000. So we have:
20,000 = 10,000 ·e0.04t
Step 3: Divide both sides by $10,000 to isolate the exponential term:
2 = e0.04t
Step 4: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.04t)
Step 5: Simplify using the properties of logarithms:
ln(2) = 0.04t·ln(e)
ln(2) = 0.04t
Step 6: Divide by 0.04 to solve for t:
t=ln(2)
0.04
Step 7: Calculate the approximate value of t:
t≈0.6931
0.04 ≈17.33
Step 8: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double in the savings account with 4
3
Question 4
Question
Samantha invested $10,000 in a savings account that pays an annual interest
rate of 5%, compounded continuously. How much money will be in the account
after 10 years?
Solution
Step 1: We can use the formula for compound interest compounded continu-
ously:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate (decimal), - tis the
time the money is invested for in years, and - eis Euler’s number, approximately
equal to 2.71828.
Step 2: Given that Samantha invests $10,000 at an annual interest rate of
5% (0.05 in decimal form), we have:
P= 10000, r = 0.05, t = 10
Plugging these values into the formula, we get:
A= 10000 ·e0.05·10
Step 3: Calculating the exponent, we have:
0.05 ·10 = 0.5
e0.5≈1.64872
Step 4: Finally, we can find the amount of money accumulated in the account
after 10 years:
A≈10000 ·1.64872 = 16487.20
Therefore, there will be approximately $16,487.20 in the account after 10
years.
Question 5
Question
A certain substance has a half-life of 3.5 hours. Suppose we start with an initial
amount of 100 grams of the substance. (a) Write an exponential decay model
for the amount of the substance remaining after thours. (b) How much of the
substance will remain after 10 hours? (c) How long will it take for the amount
of the substance to reduce to 25 grams?
4
Solution
(a) Let A(t)represent the amount of the substance remaining after thours. The
exponential decay model is given by:
A(t) = 100 ·(1
2)
t
3.5
(b) To find the remaining amount after 10 hours, we substitute t= 10 into
the model:
A(10) = 100 ·(1
2)10
3.5
A(10) = 100 ·(1
2)2.8571
A(10) ≈100 ·0.0786
A(10) ≈7.86 grams
(c) To find the time it takes for the amount to reduce to 25 grams, we set
A(t)to 25 and solve for t:
25 = 100 ·(1
2)
t
3.5
(1
2)
t
3.5
=1
4
t
3.5= 2
t= 7 ×3.5
t= 24.5hours
Question 6
Question
Solve the exponential equation: 3x−1= 27.
Solution
Step 1: Rewrite the equation in terms of a common base.
3x−1= 27
3x−1= 33
5
Step 2: Set the exponents equal to each other.
x−1 = 3
Step 3: Solve for x.
x= 3 + 1
x= 4
Step 4: Check the solution.
34−1= 27
33= 27
27 = 27
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 7
Question
Let P(t) = 5e0.08trepresent the population of a town tyears after the year 2021.
Determine the year in which the population of the town is projected to reach
50,000.
Solution
Step 1: Set up the equation to find the year when the population is 50,000. Let
P(t) = 50,000 and solve for t.
5e0.08t= 50,000
Step 2: Divide both sides by 5 to isolate the exponential term.
e0.08t= 10,000
Step 3: Take the natural logarithm of both sides to solve for t.
ln(e0.08t)= ln(10,000)
0.08t= ln(10,000)
Step 4: Simplify the right side by using the property that ln(ab)=bln(a).
0.08t= 4 ln(10)
Step 5: Divide by 0.08 to solve for t.
t=4 ln(10)
0.08
6
Step 6: Calculate the value of t.
t=4×2.3026
0.08
t≈9.2104
0.08
t≈115.255
Step 7: Since trepresents the number of years from 2021, we add 2021 to
find the year when the population is projected to reach 50,000.
2021 + 115.255 ≈2136.255
Therefore, the population of the town is projected to reach 50,000 around
the year 2136.
Question 8
Question
Suppose a bacteria culture starts with 100 bacteria and triples every hour.
1. Find a formula for the number of bacteria after thours.
2. How many bacteria will there be after 5 hours?
Solution
1. Let P(t)represent the population of bacteria after thours. Since the
number of bacteria triples every hour, we have:
P(t) = 100 ×3t
Step 1: The formula for the number of bacteria after thours is P(t) =
100 ×3t.
2. To find the number of bacteria after 5 hours, we substitute t= 5 into the
formula:
P(5) = 100 ×35= 100 ×243 = 24,300
So, after 5 hours, there will be 24,300 bacteria.
Step 2: Substituting t= 5 into the formula, we find P(5) = 24,300.
Question 9
Question
The population of a town is currently 10,000 and is expected to triple every 20
years. Write an exponential function to model the population of the town after
tyears. What will be the population of the town in 60 years?
7
Solution
Step 1: Let P(t)be the population of the town after tyears. Since the population
is tripling every 20 years, the exponential model is given by P(t) = 10,000 ×
3t/20.
Step 2: To find the population of the town in 60 years, substitute t= 60
into the formula:
P(60) = 10,000 ×360/20
P(60) = 10,000 ×33
P(60) = 10,000 ×27
P(60) = 270,000
Therefore, the population of the town after 60 years will be 270,000.
Question 10
Question
Samantha deposits $5,000 into a savings account that pays 3.5% interest com-
pounded continuously. How much will be in the account after 10 years?
Solution
Step 1: We can use the formula for compound interest compounded continu-
ously:
A=P ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial deposit), r= annual interest rate (decimal), t= time in years,
and eis the base of the natural logarithm (approximately 2.71828).
Step 2: Given that: P= $5,000,r= 0.035 (3.5% expressed as a decimal),
and t= 10 years.
Step 3: Substituting the given values into the formula, we get:
A= 5000 ·e0.035·10
Step 4: Calculating the exponent:
A= 5000 ·e0.35
Step 5: Evaluating the exponential term:
A≈5000 ·1.419067
Step 6: Finally, we find the amount accumulated after 10 years:
A≈$7,095.34
Therefore, after 10 years, there will be approximately $7,095.34 in the ac-
count.
8
Question 11
Question
Solve for x:32x+1 = 54
Solution
Step 1: Rewrite 54 as a power of 3 by recognizing that 54 = 33.
32x+1 = 33
Step 2: Set the exponents equal, since the bases are the same.
2x+ 1 = 3
Step 3: Solve for x. Subtract 1 from both sides:
2x= 2
Divide by 2:
x= 1
Step 4: Therefore, the solution to the equation 32x+1 = 54 is x= 1 .
Question 12
Question
Solve the following logarithmic equation for x:3 log(x+ 2) −2 log(x) = 4.
Solution
Step 1: Use the properties of logarithms to simplify the equation. We know
that log(a)−log(b) = log (a
band log(an) = nlog(a). Applying these properties,
we get:
3 log(x+ 2) −2 log(x) = log((x+ 2)3)−log(x2)
Step 2: Simplify the equation further:
log((x+ 2)3)−log(x2)= 4
log ((x+ 2)3
x2)= 4
Step 3: Rewrite the equation in exponential form. We know that logb(a) = c
is equivalent to bc=a. Applying this to our equation, we have:
(x+ 2)3
x2= 104
9
Step 4: Simplify the right side:
(x+ 2)3= 104·x2
(x+ 2)3= 10000x2
Step 5: Expand the left side of the equation:
x3+ 6x2+ 12x+ 8 = 10000x2
Step 6: Rearrange the equation and set it equal to zero:
x3−9994x2+ 12x+ 8 = 0
Step 7: This equation doesn’t seem to simplify easily. To solve this equation,
we may need to use numerical methods or a calculator.
Therefore, the solution to the given logarithmic equation is xvalues that
satisfy the equation x3−9994x2+ 12x+ 8 = 0.
Question 13
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since the population was first measured.
Calculate the population of the city after 10 years.
Solution
Step 1: Substitute t= 10 into the population model:
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210:
1.0210 = 1.218391
Step 3: Substitute this value back into the population model to find P(10):
P(10) = 5000 ·1.218391 = 6091.955
Step 4: Therefore, after 10 years, the population of the city would be ap-
proximately 6092 people.
Question 14
Question
Samantha deposited $5000 into a savings account that earns 3.8% annual inter-
est, compounded quarterly. How much will be in the account after 10 years?
10
Solution
Step 1: Calculate the interest rate per compounding period. Given that the
annual interest rate is 3.8%, the quarterly interest rate is 3.8%/4=0.95% =
0.0095 in decimal form.
Step 2: Determine the number of compounding periods over 10 years. Since
the interest is compounded quarterly, there are 4compounding periods per year
and 4×10 = 40 total compounding periods over 10 years.
Step 3: Use the compound interest formula to find the future value of the
account. The formula for compound interest is:
A=P(1 + r
n)nt
where: - Ais the future value of the account, - Pis the principal amount ($5000
in this case), - ris the interest rate per compounding period (0.0095), - nis the
number of compounding periods per year (4), - and tis the number of years the
money is invested (10).
Substitute the values into the formula:
A= 5000 (1 + 0.0095
4)4×10
Step 4: Perform the calculations to find the future value of the account.
A= 5000 (1 + 0.002375)40
A= 5000 ×1.00237540
A≈5000 ×1.4141
A≈$7070.50
Therefore, after 10 years, there will be approximately $7070.50 in the ac-
count.
Question 15
Question
A certain population of bacteria triples every 5 hours. If there are initially 100
bacteria, how many bacteria will be present after 15 hours?
Solution
Step 1: Write down the formula for exponential growth. The formula for expo-
nential growth is given by:
N(t) = N0·at/d,
11
where: N(t)is the population at time t,N0is the initial population, ais the
growth factor, tis time elapsed, and dis the time it takes for the population to
grow by a factor of a.
Step 2: Find the growth factor. Since the population triples every 5 hours,
the growth factor ais 3, because 31= 3.
Step 3: Substitute the given values into the formula. Substitute N0= 100,
a= 3, and t= 15 into the formula to find N(15):
N(15) = 100 ·315
5.
Step 4: Calculate the population at 15 hours.
N(15) = 100 ·33= 100 ·27 = 2700.
Therefore, after 15 hours, there will be 2700 bacteria present.
Question 16
Question
Solve the following logarithmic equation for x:log2(3x+ 4) = log2(x+ 7).
Solution
Step 1: Since both sides of the equation are logarithms with base 2, we can
eliminate the logarithms by setting the arguments equal to each other:
3x+ 4 = x+ 7
Step 2: Next, we solve this linear equation for x:
3x+ 4 = x+ 7
2x= 3
x=3
2
Step 3: Therefore, the solution to the logarithmic equation log2(3x+ 4) =
log2(x+ 7) is x=3
2.
Question 17
Question
Suppose a researcher is studying the growth of a certain bacteria culture in a
controlled environment. The researcher determines that the bacteria population
doubles every 4 hours. If the initial population is 100 bacteria, what is the
exponential equation that models the population of the bacteria as a function
of time?
12
Solution
Step 1: Let’s denote the initial bacteria population as P0= 100 bacteria and
the time in hours as t. Since the population doubles every 4 hours, the growth
rate is 2.
Step 2: The exponential equation that models the population of the bacteria
as a function of time can be expressed as P(t) = P0·2t/4.
Step 3: Substitute P0= 100 into the equation to get P(t) = 100 ·2t/4.
Step 4: Therefore, the exponential equation that models the population of
the bacteria as a function of time is P(t) = 100 ·2t/4.
Question 18
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.03t, where
tis the number of years since the year 2020. Find the population of the city in
the year 2030.
Solution
Step 1: To find the population in the year 2030, we need to substitute t= 10
into the population function P(t). Step 2: Substitute t= 10 into the population
function:
P(10) = 5000 ·e0.03·10
Step 3: Simplify the expression:
P(10) = 5000 ·e0.3
Step 4: Use the fact that e≈2.71828 to approximate the population in the year
2030:
P(10) = 5000 ·2.718280.3
Step 5: Calculate the approximate population of the city in the year 2030:
P(10) ≈5000 ·1.349858
P(10) ≈6749.29
Step 6: Therefore, the population of the city in the year 2030 is approximately
6,749.29.
Question 19
Question
A certain radioactive substance decays according to the formula A(t) = A0e−kt,
where A(t)is the amount of substance remaining after tyears, A0is the initial
amount of substance, and kis a positive constant. If 80
13
Solution
Step 1: We are given that 80
0.20A0=A0e−10k
Dividing both sides by A0, we get:
0.20 = e−10k
Taking the natural logarithm of both sides, we have:
ln(0.20) = ln(e−10k)
ln(0.20) = −10k
k=−ln(0.20)
10
Step 2: Now, we want to find out how long it will take for 95
0.05A0=A0e−kt
0.05 = e−(−ln(0.20)
10 )t
0.05 = eln(0.20)
10 t
Step 3: To solve for t, take the natural logarithm of both sides:
ln(0.05) = ln(eln(0.20)
10 t)
ln(0.05) = ln(0.20)
10 t
t=10 ln(0.05)
ln(0.20)
Therefore, it will take approximately 10 ln(0.05)
ln(0.20) years for 95
Question 20
Question
The population of a city is modeled by the function P(t) = 5000 ·1.03t, where
P(t)represents the population after tyears. How long will it take for the
population to reach 10,000?
14
Solution
Step 1: Set up the equation by substituting P(t) = 10000 into the population
function:
10000 = 5000 ·1.03t
Step 2: Divide both sides by 5000 to isolate the exponential term:
10000
5000 = 1.03t
Step 3: Simplify the left side:
2 = 1.03t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(1.03t)
Step 5: Apply the properties of logarithms to bring down the exponent:
ln(2) = t·ln(1.03)
Step 6: Divide both sides by ln(1.03) to solve for t:
t=ln(2)
ln(1.03)
Step 7: Calculate the approximate value of tusing a calculator:
t≈ln(2)
ln(1.03) ≈22.44
So, it will take approximately 22.44 years for the population to reach 10,000
in the city.
Question 21
Question
Samantha has invested $5000 in a savings account that gives her an annual
interest rate of 5%, compounded continuously. How much will Samantha have
after 10 years?
Solution
Step 1: We can use the formula for continuous compound interest:
A=P·ert
15
where: - Ais the amount of money accumulated after tyears, - Pis the initial
investment ($5000 in this case), - ris the annual interest rate (expressed as a
decimal, so r= 0.05 in this case), and - tis the number of years the money is
invested for.
Step 2: Plug in the values into the formula to find the amount of money
Samantha will have after 10 years:
A= 5000 ·e0.05·10
Step 3: Calculate the value of the exponent:
0.05 ·10 = 0.5
Step 4: Substitute the exponent back into the formula:
A= 5000 ·e0.5
Step 5: Calculate the final amount by evaluating the exponential function:
A= 5000 ·e0.5≈5000 ·1.648721 ≈8243.61
Therefore, after 10 years, Samantha will have approximately $8243.61 in her
savings account.
Question 22
Question
The population of a city is growing exponentially with a growth rate of 3.5
Solution
Step 1: Determine the growth factor. Let P(t)represent the population at time
t, and rbe the annual growth rate in decimal form. The growth factor is given
by the formula:
F= 1 + r
In this case, r= 0.035, so the growth factor is F= 1 + 0.035 = 1.035.
Step 2: Find the population after 10 years. The population after tyears can
be modeled by the formula:
P(t) = P0·Ft
where P0is the initial population. Given that the current population is P0=
200,000 and we want to find the population in 10 years, we substitute t= 10
into the formula:
P(10) = 200,000 ·1.03510
Step 3: Calculate the population after 10 years. Now, we compute:
P(10) = 200,000 ·1.03510 ≈200,000 ·1.424 = 284,800
So, the estimated population of the city after 10 years will be approximately
284,800.
16
Question 23
Question
Solve the exponential equation 3x+1 = 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3. Since 27 = 33, the equation becomes
3x+1 = 33.
Step 2: Set the exponents equal to each other. We have x+ 1 = 3.
Step 3: Solve for xby subtracting 1 from both sides. We get x= 3 −1.
Step 4: Simplify to find the final answer. Therefore, x= 2.
Thus, the solution to the exponential equation 3x+1 = 27 is x= 2.
Question 24
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the exponents equal to each other.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
Step 4: Solve for x.
2x= 3 −1
2x= 2
x= 1
Step 5: Check the solution.
32(1)+1 = 33
33= 27
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
17
Question 25
Question
Sara invested $5000 in a savings account that earns 2% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Let trepresent the time it takes for the investment to double in value. To solve
for t, we can use the continuous compound interest formula:
A=P·ert
Where: - Ais the amount of money after time t, - Pis the initial investment
amount, - ris the interest rate, - eis the base of the natural logarithm, - tis
the time.
In this case, Sara’s initial investment is $5000, the interest rate is 2% = 0.02,
and the amount she wants to reach is 2·$5000 = $10000.
Step 1: Set up the equation using the given information:
$10000 = $5000 ·e0.02t
Step 2: Divide both sides by $5000 to isolate e0.02t:
2 = e0.02t
Step 3: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.02t)
Step 4: Use the property of logarithms to simplify:
ln 2 = 0.02t·ln(e)
ln 2 = 0.02t
Step 5: Finally, solve for tby dividing both sides by 0.02:
t=ln 2
0.02
Thus, it will take approximately 34.66 years for Sara’s investment to double
in value.
Question 26
Question
The population of a city is modeled by the function P(t) = 25000 ·1.02t, where
trepresents the number of years since 2020. Calculate the population of the
city in 2040 according to this model.
18
Solution
Step 1: Determine the value of tin 2040. Since 2040 is 20 years after 2020,
t= 20.
Step 2: Substitute t= 20 into the population function.
P(20) = 25000 ·1.0220
Step 3: Calculate the population in 2040.
P(20) = 25000 ·1.0220 = 25000 ·2.191 = 54775
Therefore, according to the model, the population of the city in 2040 will be
54,775.
Question 27
Question
Solve the exponential equation for x:3x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: Since 27 can be written as 33, we have:
3x+1 = 33
Step 2: Now, we can set the exponents equal to each other:
x+ 1 = 3
Solving for x, we get
x= 3 −1
x= 2
Therefore, the solution to the equation is x= 2.
Question 28
Question
Solve the exponential equation 32x+1 = 27.
19
Solution
Step 1: Rewrite the equation in terms of a common base. Step 2: Solve for x.
Step 1: Rewrite the equation in terms of a common base. Since 27 = 33,
we can rewrite the equation 32x+1 = 27 as 32x+1 = 33.
Step 2: Solve for x. Since the bases are the same, we can set the exponents
equal to each other:
2x+ 1 = 3
Subtract 1 from both sides:
2x= 2
Divide by 2:
x= 1
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 29
Question
Solve the exponential equation 3x−1= 9.
Solution
Step 1: Rewrite 9as a power of 3.
3x−1= 32
Step 2: Since the bases are the same, set the exponents equal to each other.
(x−1) = 2
Step 3: Solve for x.
x= 2 + 1
Step 4: Simplify.
x= 3
Therefore, the solution to the exponential equation 3x−1= 9 is x= 3.
Question 30
Question
Solve the following exponential equation for x:
(5x) + (5x−1) = 150.
20
Solution
Step 1: Recall the properties of exponents to simplify the equation. Step 2:
Rewrite the equation in terms of a single base. Step 3: Solve the resulting
equation for x.
Step 1: Simplify the equation using the properties of exponents.
We know that am·an=am+n, so we can rewrite the given equation as:
5x+ 5x−1= 150.
Step 2: Rewrite the equation in terms of a single base.
We know that 5x−1=5x
5. Substituting this into the equation, we get:
5x+5x
5= 150.
Now, we can combine the terms:
5x+1
5·5x= 150.
Step 3: Solve the resulting equation for x.
Combining like terms, we have:
6
5·5x= 150.
Dividing by 6
5on both sides, we get:
5x=150 ·5
6= 125.
Taking the logarithm of both sides to solve for x, we have:
log5(5x) = log5(125).
This implies:
x= log5(125) = 3.
Therefore, the solution to the equation (5x) + (5x−1) = 150 is x= 3.
Question 31
Question
Solve the exponential equation 42x−1= 32.
21
Solution
Step 1: Rewrite 32 as a power of 4by noting that 32 = 42·2.
42x−1= 42·2
Step 2: Simplify the equation by equating the exponents.
2x−1 = 2 + 1
Step 3: Solve for xby isolating it.
2x= 3 + 1
2x= 4
Step 4: Divide by 2to solve for x.
x=4
2
Step 5: Simplify to find the final solution for x.
x= 2
Therefore, the solution to the exponential equation 42x−1= 32 is x= 2.
Question 32
Question
Solve the following exponential equation for x:3x+ 3x−1= 20.
Solution
Step 1: Rewrite 3x−1=1
3·3xto get the equation in terms of a single exponential
term. Step 2: Substitute 3x−1with 1
3·3xin the original equation to get 3x+
1
3·3x= 20. Step 3: Combine like terms to simplify the equation to 4
3·3x= 20.
Step 4: Multiply both sides by 3
4to isolate 3xon one side: 3x=15
2. Step 5:
Rewrite 15
2as 3log315
2. Step 6: Equate the exponents to solve for x:x= log315
2.
Step 7: Use a calculator to approximate x≈1.861.
Question 33
Question
Solve the exponential equation 3x−1+ 32x−1= 28.
22
Solution
Step 1: Let’s first rewrite 3x−1+ 32x−1= 28 in terms of the same base. We
can rewrite 32x−1as (32)x·3−1= 9x·1
3. Therefore, our equation becomes
3x−1+ 9x·1
3= 28.
Step 2: Multiply the entire equation by 3 to get rid of the fraction, resulting
in 3·3x−1+ 9x= 84.
Step 3: Utilize the properties of exponents to simplify the equation further.
3·3x−1= 31·3x−1= 3x. Therefore, the equation becomes 3x+ 9x= 84.
Step 4: Recognize that 9 = 32, so we can rewrite 9xas (32)x= 32x. Substi-
tuting this back into the equation gives us 3x+ 32x= 84.
Step 5: Factor out 3xfrom both terms to get 3x(1 + 3x) = 84.
Step 6: Now we can solve for 3xby dividing both sides by (1 + 3x), which
gives 3x=84
1+3x.
Step 7: Substitute y= 3x, which transforms the equation into y=84
1+y.
Step 8: Rearrange the equation to solve for y:y(1 + y) = 84. Simplify to
get y2+y−84 = 0.
Step 9: Factor the quadratic equation to get (y+ 12)(y−7) = 0. Therefore,
y=−12 or y= 7.
Step 10: Restore yto 3xto find that 3x=−12 does not have a real solution.
However, when 3x= 7, we find one valid solution: x= log37.
Question 34
Question
Solve the equation e2x= 10 for x. Round your answer to four decimal places if
necessary.
Solution
Step 1: Take the natural logarithm of both sides of the equation to eliminate
the base e.
ln(e2x)= ln(10)
Step 2: Use the property of logarithms that ln(ab)=bln(a).
2xln(e) = ln(10)
Step 3: Since ln(e) = 1, we have:
2x= ln(10)
Step 4: Divide by 2 to solve for x.
x=ln(10)
2≈0.6931
Therefore, the solution to the equation e2x= 10 is approximately x≈0.6931.
23
Question 35
Question
Samantha invested $10,000 in a savings account that earns 4.5% interest com-
pounded continuously. How long will it take for her investment to double in
value?
Solution
Let Arepresent the amount of money in the account at any time t, measured
in years. We know that the formula for continuous compound interest is given
by A=P ert, where: - Pis the initial principal amount ($10,000 in this case), -
ris the annual interest rate in decimal form (4.5% or 0.045 in this case), - tis
the time in years, and - eis the constant approximately equal to 2.71828.
We want to find the time it takes for the investment to double, so we are
looking for twhen A= 2P.
Step 1: Set up the equation for A:
A= 10000e0.045t
Step 2: Set Ato be double the initial amount:
2×10000 = 10000e0.045t
Step 3: Simplify the equation and solve for t:
20000 = 10000e0.045t
2 = e0.045t
Step 4: Take natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
ln(2) = 0.045t
t=ln(2)
0.045
Step 5: Calculate the value of t:
t≈ln(2)
0.045 ≈15.40 years
Therefore, it will take approximately 15.40 years for Samantha’s investment
to double in value when earning 4.5% interest compounded continuously.
24
Question 2
Question
Solve for x:23x+1 −5·2x+1 + 6 = 0.
Solution
Step 1: Let y= 2x. Then, rewrite the equation using y.
23x+1 −5·2x+1 + 6 = 0
⇒2·23x−5·2·2x+ 6 = 0
⇒2·(2x)3−5·2·2x+ 6 = 0
⇒2y3−10y+ 6 = 0
Step 2: Solve the cubic equation by factoring or using any appropriate
method.
Since the coefficient of y3is 2we can simplify the equation by dividing by
2:
y3−5y+ 3 = 0
Step 3: Try potential roots by using the Rational Root Theorem. The
possible rational roots are ±1,±3.
Substitute these values to find the root. We find y= 1 is a root by trial and
error.
Step 4: Use synthetic division to divide y3−5y+ 3 by (y−1).
1 1 0 −5
3
1 1
−4
Step 5: The synthetic division shows that y3−5y+ 3 = (y−1)(y2+y−3).
Step 6: Factor the quadratic polynomial y2+y−3.
y2+y−3 = (y+ 2)(y−1)
Step 7: Set each factor equal to zero and solve for y.
y+ 2 = 0 ⇒y=−2
y−1 = 0 ⇒y= 1
Step 8: Substitute back y= 2xto find the corresponding values of x:
y= 1 ⇒2x= 1 ⇒x= 0
y=−2⇒2x=−2(No real solution)
Step 9: Therefore, the solution to the equation 23x+1 −5·2x+1 + 6 = 0 is
x= 0.
2
Question 3
Question
Samantha invested $10,000 in a savings account that earns 4
Solution
Step 1: The formula for continuously compounded interest is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (initial investment). - ris the annual interest rate
(in decimal form). - tis the time the money is invested for in years. - eis Euler’s
number, approximately equal to 2.71828.
Step 2: Since Samantha wants to double her investment, the amount after t
years will be $20,000. So we have:
20,000 = 10,000 ·e0.04t
Step 3: Divide both sides by $10,000 to isolate the exponential term:
2 = e0.04t
Step 4: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.04t)
Step 5: Simplify using the properties of logarithms:
ln(2) = 0.04t·ln(e)
ln(2) = 0.04t
Step 6: Divide by 0.04 to solve for t:
t=ln(2)
0.04
Step 7: Calculate the approximate value of t:
t≈0.6931
0.04 ≈17.33
Step 8: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double in the savings account with 4
3
Question 4
Question
Samantha invested $10,000 in a savings account that pays an annual interest
rate of 5%, compounded continuously. How much money will be in the account
after 10 years?
Solution
Step 1: We can use the formula for compound interest compounded continu-
ously:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate (decimal), - tis the
time the money is invested for in years, and - eis Euler’s number, approximately
equal to 2.71828.
Step 2: Given that Samantha invests $10,000 at an annual interest rate of
5% (0.05 in decimal form), we have:
P= 10000, r = 0.05, t = 10
Plugging these values into the formula, we get:
A= 10000 ·e0.05·10
Step 3: Calculating the exponent, we have:
0.05 ·10 = 0.5
e0.5≈1.64872
Step 4: Finally, we can find the amount of money accumulated in the account
after 10 years:
A≈10000 ·1.64872 = 16487.20
Therefore, there will be approximately $16,487.20 in the account after 10
years.
Question 5
Question
A certain substance has a half-life of 3.5 hours. Suppose we start with an initial
amount of 100 grams of the substance. (a) Write an exponential decay model
for the amount of the substance remaining after thours. (b) How much of the
substance will remain after 10 hours? (c) How long will it take for the amount
of the substance to reduce to 25 grams?
4
Solution
(a) Let A(t)represent the amount of the substance remaining after thours. The
exponential decay model is given by:
A(t) = 100 ·(1
2)
t
3.5
(b) To find the remaining amount after 10 hours, we substitute t= 10 into
the model:
A(10) = 100 ·(1
2)10
3.5
A(10) = 100 ·(1
2)2.8571
A(10) ≈100 ·0.0786
A(10) ≈7.86 grams
(c) To find the time it takes for the amount to reduce to 25 grams, we set
A(t)to 25 and solve for t:
25 = 100 ·(1
2)
t
3.5
(1
2)
t
3.5
=1
4
t
3.5= 2
t= 7 ×3.5
t= 24.5hours
Question 6
Question
Solve the exponential equation: 3x−1= 27.
Solution
Step 1: Rewrite the equation in terms of a common base.
3x−1= 27
3x−1= 33
5
Step 2: Set the exponents equal to each other.
x−1 = 3
Step 3: Solve for x.
x= 3 + 1
x= 4
Step 4: Check the solution.
34−1= 27
33= 27
27 = 27
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 7
Question
Let P(t) = 5e0.08trepresent the population of a town tyears after the year 2021.
Determine the year in which the population of the town is projected to reach
50,000.
Solution
Step 1: Set up the equation to find the year when the population is 50,000. Let
P(t) = 50,000 and solve for t.
5e0.08t= 50,000
Step 2: Divide both sides by 5 to isolate the exponential term.
e0.08t= 10,000
Step 3: Take the natural logarithm of both sides to solve for t.
ln(e0.08t)= ln(10,000)
0.08t= ln(10,000)
Step 4: Simplify the right side by using the property that ln(ab)=bln(a).
0.08t= 4 ln(10)
Step 5: Divide by 0.08 to solve for t.
t=4 ln(10)
0.08
6
Step 6: Calculate the value of t.
t=4×2.3026
0.08
t≈9.2104
0.08
t≈115.255
Step 7: Since trepresents the number of years from 2021, we add 2021 to
find the year when the population is projected to reach 50,000.
2021 + 115.255 ≈2136.255
Therefore, the population of the town is projected to reach 50,000 around
the year 2136.
Question 8
Question
Suppose a bacteria culture starts with 100 bacteria and triples every hour.
1. Find a formula for the number of bacteria after thours.
2. How many bacteria will there be after 5 hours?
Solution
1. Let P(t)represent the population of bacteria after thours. Since the
number of bacteria triples every hour, we have:
P(t) = 100 ×3t
Step 1: The formula for the number of bacteria after thours is P(t) =
100 ×3t.
2. To find the number of bacteria after 5 hours, we substitute t= 5 into the
formula:
P(5) = 100 ×35= 100 ×243 = 24,300
So, after 5 hours, there will be 24,300 bacteria.
Step 2: Substituting t= 5 into the formula, we find P(5) = 24,300.
Question 9
Question
The population of a town is currently 10,000 and is expected to triple every 20
years. Write an exponential function to model the population of the town after
tyears. What will be the population of the town in 60 years?
7
Solution
Step 1: Let P(t)be the population of the town after tyears. Since the population
is tripling every 20 years, the exponential model is given by P(t) = 10,000 ×
3t/20.
Step 2: To find the population of the town in 60 years, substitute t= 60
into the formula:
P(60) = 10,000 ×360/20
P(60) = 10,000 ×33
P(60) = 10,000 ×27
P(60) = 270,000
Therefore, the population of the town after 60 years will be 270,000.
Question 10
Question
Samantha deposits $5,000 into a savings account that pays 3.5% interest com-
pounded continuously. How much will be in the account after 10 years?
Solution
Step 1: We can use the formula for compound interest compounded continu-
ously:
A=P ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial deposit), r= annual interest rate (decimal), t= time in years,
and eis the base of the natural logarithm (approximately 2.71828).
Step 2: Given that: P= $5,000,r= 0.035 (3.5% expressed as a decimal),
and t= 10 years.
Step 3: Substituting the given values into the formula, we get:
A= 5000 ·e0.035·10
Step 4: Calculating the exponent:
A= 5000 ·e0.35
Step 5: Evaluating the exponential term:
A≈5000 ·1.419067
Step 6: Finally, we find the amount accumulated after 10 years:
A≈$7,095.34
Therefore, after 10 years, there will be approximately $7,095.34 in the ac-
count.
8
Question 11
Question
Solve for x:32x+1 = 54
Solution
Step 1: Rewrite 54 as a power of 3 by recognizing that 54 = 33.
32x+1 = 33
Step 2: Set the exponents equal, since the bases are the same.
2x+ 1 = 3
Step 3: Solve for x. Subtract 1 from both sides:
2x= 2
Divide by 2:
x= 1
Step 4: Therefore, the solution to the equation 32x+1 = 54 is x= 1 .
Question 12
Question
Solve the following logarithmic equation for x:3 log(x+ 2) −2 log(x) = 4.
Solution
Step 1: Use the properties of logarithms to simplify the equation. We know
that log(a)−log(b) = log (a
band log(an) = nlog(a). Applying these properties,
we get:
3 log(x+ 2) −2 log(x) = log((x+ 2)3)−log(x2)
Step 2: Simplify the equation further:
log((x+ 2)3)−log(x2)= 4
log ((x+ 2)3
x2)= 4
Step 3: Rewrite the equation in exponential form. We know that logb(a) = c
is equivalent to bc=a. Applying this to our equation, we have:
(x+ 2)3
x2= 104
9
Step 4: Simplify the right side:
(x+ 2)3= 104·x2
(x+ 2)3= 10000x2
Step 5: Expand the left side of the equation:
x3+ 6x2+ 12x+ 8 = 10000x2
Step 6: Rearrange the equation and set it equal to zero:
x3−9994x2+ 12x+ 8 = 0
Step 7: This equation doesn’t seem to simplify easily. To solve this equation,
we may need to use numerical methods or a calculator.
Therefore, the solution to the given logarithmic equation is xvalues that
satisfy the equation x3−9994x2+ 12x+ 8 = 0.
Question 13
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since the population was first measured.
Calculate the population of the city after 10 years.
Solution
Step 1: Substitute t= 10 into the population model:
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210:
1.0210 = 1.218391
Step 3: Substitute this value back into the population model to find P(10):
P(10) = 5000 ·1.218391 = 6091.955
Step 4: Therefore, after 10 years, the population of the city would be ap-
proximately 6092 people.
Question 14
Question
Samantha deposited $5000 into a savings account that earns 3.8% annual inter-
est, compounded quarterly. How much will be in the account after 10 years?
10
Solution
Step 1: Calculate the interest rate per compounding period. Given that the
annual interest rate is 3.8%, the quarterly interest rate is 3.8%/4=0.95% =
0.0095 in decimal form.
Step 2: Determine the number of compounding periods over 10 years. Since
the interest is compounded quarterly, there are 4compounding periods per year
and 4×10 = 40 total compounding periods over 10 years.
Step 3: Use the compound interest formula to find the future value of the
account. The formula for compound interest is:
A=P(1 + r
n)nt
where: - Ais the future value of the account, - Pis the principal amount ($5000
in this case), - ris the interest rate per compounding period (0.0095), - nis the
number of compounding periods per year (4), - and tis the number of years the
money is invested (10).
Substitute the values into the formula:
A= 5000 (1 + 0.0095
4)4×10
Step 4: Perform the calculations to find the future value of the account.
A= 5000 (1 + 0.002375)40
A= 5000 ×1.00237540
A≈5000 ×1.4141
A≈$7070.50
Therefore, after 10 years, there will be approximately $7070.50 in the ac-
count.
Question 15
Question
A certain population of bacteria triples every 5 hours. If there are initially 100
bacteria, how many bacteria will be present after 15 hours?
Solution
Step 1: Write down the formula for exponential growth. The formula for expo-
nential growth is given by:
N(t) = N0·at/d,
11
where: N(t)is the population at time t,N0is the initial population, ais the
growth factor, tis time elapsed, and dis the time it takes for the population to
grow by a factor of a.
Step 2: Find the growth factor. Since the population triples every 5 hours,
the growth factor ais 3, because 31= 3.
Step 3: Substitute the given values into the formula. Substitute N0= 100,
a= 3, and t= 15 into the formula to find N(15):
N(15) = 100 ·315
5.
Step 4: Calculate the population at 15 hours.
N(15) = 100 ·33= 100 ·27 = 2700.
Therefore, after 15 hours, there will be 2700 bacteria present.
Question 16
Question
Solve the following logarithmic equation for x:log2(3x+ 4) = log2(x+ 7).
Solution
Step 1: Since both sides of the equation are logarithms with base 2, we can
eliminate the logarithms by setting the arguments equal to each other:
3x+ 4 = x+ 7
Step 2: Next, we solve this linear equation for x:
3x+ 4 = x+ 7
2x= 3
x=3
2
Step 3: Therefore, the solution to the logarithmic equation log2(3x+ 4) =
log2(x+ 7) is x=3
2.
Question 17
Question
Suppose a researcher is studying the growth of a certain bacteria culture in a
controlled environment. The researcher determines that the bacteria population
doubles every 4 hours. If the initial population is 100 bacteria, what is the
exponential equation that models the population of the bacteria as a function
of time?
12
Solution
Step 1: Let’s denote the initial bacteria population as P0= 100 bacteria and
the time in hours as t. Since the population doubles every 4 hours, the growth
rate is 2.
Step 2: The exponential equation that models the population of the bacteria
as a function of time can be expressed as P(t) = P0·2t/4.
Step 3: Substitute P0= 100 into the equation to get P(t) = 100 ·2t/4.
Step 4: Therefore, the exponential equation that models the population of
the bacteria as a function of time is P(t) = 100 ·2t/4.
Question 18
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.03t, where
tis the number of years since the year 2020. Find the population of the city in
the year 2030.
Solution
Step 1: To find the population in the year 2030, we need to substitute t= 10
into the population function P(t). Step 2: Substitute t= 10 into the population
function:
P(10) = 5000 ·e0.03·10
Step 3: Simplify the expression:
P(10) = 5000 ·e0.3
Step 4: Use the fact that e≈2.71828 to approximate the population in the year
2030:
P(10) = 5000 ·2.718280.3
Step 5: Calculate the approximate population of the city in the year 2030:
P(10) ≈5000 ·1.349858
P(10) ≈6749.29
Step 6: Therefore, the population of the city in the year 2030 is approximately
6,749.29.
Question 19
Question
A certain radioactive substance decays according to the formula A(t) = A0e−kt,
where A(t)is the amount of substance remaining after tyears, A0is the initial
amount of substance, and kis a positive constant. If 80
13
Solution
Step 1: We are given that 80
0.20A0=A0e−10k
Dividing both sides by A0, we get:
0.20 = e−10k
Taking the natural logarithm of both sides, we have:
ln(0.20) = ln(e−10k)
ln(0.20) = −10k
k=−ln(0.20)
10
Step 2: Now, we want to find out how long it will take for 95
0.05A0=A0e−kt
0.05 = e−(−ln(0.20)
10 )t
0.05 = eln(0.20)
10 t
Step 3: To solve for t, take the natural logarithm of both sides:
ln(0.05) = ln(eln(0.20)
10 t)
ln(0.05) = ln(0.20)
10 t
t=10 ln(0.05)
ln(0.20)
Therefore, it will take approximately 10 ln(0.05)
ln(0.20) years for 95
Question 20
Question
The population of a city is modeled by the function P(t) = 5000 ·1.03t, where
P(t)represents the population after tyears. How long will it take for the
population to reach 10,000?
14
Solution
Step 1: Set up the equation by substituting P(t) = 10000 into the population
function:
10000 = 5000 ·1.03t
Step 2: Divide both sides by 5000 to isolate the exponential term:
10000
5000 = 1.03t
Step 3: Simplify the left side:
2 = 1.03t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(1.03t)
Step 5: Apply the properties of logarithms to bring down the exponent:
ln(2) = t·ln(1.03)
Step 6: Divide both sides by ln(1.03) to solve for t:
t=ln(2)
ln(1.03)
Step 7: Calculate the approximate value of tusing a calculator:
t≈ln(2)
ln(1.03) ≈22.44
So, it will take approximately 22.44 years for the population to reach 10,000
in the city.
Question 21
Question
Samantha has invested $5000 in a savings account that gives her an annual
interest rate of 5%, compounded continuously. How much will Samantha have
after 10 years?
Solution
Step 1: We can use the formula for continuous compound interest:
A=P·ert
15
where: - Ais the amount of money accumulated after tyears, - Pis the initial
investment ($5000 in this case), - ris the annual interest rate (expressed as a
decimal, so r= 0.05 in this case), and - tis the number of years the money is
invested for.
Step 2: Plug in the values into the formula to find the amount of money
Samantha will have after 10 years:
A= 5000 ·e0.05·10
Step 3: Calculate the value of the exponent:
0.05 ·10 = 0.5
Step 4: Substitute the exponent back into the formula:
A= 5000 ·e0.5
Step 5: Calculate the final amount by evaluating the exponential function:
A= 5000 ·e0.5≈5000 ·1.648721 ≈8243.61
Therefore, after 10 years, Samantha will have approximately $8243.61 in her
savings account.
Question 22
Question
The population of a city is growing exponentially with a growth rate of 3.5
Solution
Step 1: Determine the growth factor. Let P(t)represent the population at time
t, and rbe the annual growth rate in decimal form. The growth factor is given
by the formula:
F= 1 + r
In this case, r= 0.035, so the growth factor is F= 1 + 0.035 = 1.035.
Step 2: Find the population after 10 years. The population after tyears can
be modeled by the formula:
P(t) = P0·Ft
where P0is the initial population. Given that the current population is P0=
200,000 and we want to find the population in 10 years, we substitute t= 10
into the formula:
P(10) = 200,000 ·1.03510
Step 3: Calculate the population after 10 years. Now, we compute:
P(10) = 200,000 ·1.03510 ≈200,000 ·1.424 = 284,800
So, the estimated population of the city after 10 years will be approximately
284,800.
16
Question 23
Question
Solve the exponential equation 3x+1 = 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3. Since 27 = 33, the equation becomes
3x+1 = 33.
Step 2: Set the exponents equal to each other. We have x+ 1 = 3.
Step 3: Solve for xby subtracting 1 from both sides. We get x= 3 −1.
Step 4: Simplify to find the final answer. Therefore, x= 2.
Thus, the solution to the exponential equation 3x+1 = 27 is x= 2.
Question 24
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the exponents equal to each other.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
Step 4: Solve for x.
2x= 3 −1
2x= 2
x= 1
Step 5: Check the solution.
32(1)+1 = 33
33= 27
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
17
Question 25
Question
Sara invested $5000 in a savings account that earns 2% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Let trepresent the time it takes for the investment to double in value. To solve
for t, we can use the continuous compound interest formula:
A=P·ert
Where: - Ais the amount of money after time t, - Pis the initial investment
amount, - ris the interest rate, - eis the base of the natural logarithm, - tis
the time.
In this case, Sara’s initial investment is $5000, the interest rate is 2% = 0.02,
and the amount she wants to reach is 2·$5000 = $10000.
Step 1: Set up the equation using the given information:
$10000 = $5000 ·e0.02t
Step 2: Divide both sides by $5000 to isolate e0.02t:
2 = e0.02t
Step 3: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.02t)
Step 4: Use the property of logarithms to simplify:
ln 2 = 0.02t·ln(e)
ln 2 = 0.02t
Step 5: Finally, solve for tby dividing both sides by 0.02:
t=ln 2
0.02
Thus, it will take approximately 34.66 years for Sara’s investment to double
in value.
Question 26
Question
The population of a city is modeled by the function P(t) = 25000 ·1.02t, where
trepresents the number of years since 2020. Calculate the population of the
city in 2040 according to this model.
18
Solution
Step 1: Determine the value of tin 2040. Since 2040 is 20 years after 2020,
t= 20.
Step 2: Substitute t= 20 into the population function.
P(20) = 25000 ·1.0220
Step 3: Calculate the population in 2040.
P(20) = 25000 ·1.0220 = 25000 ·2.191 = 54775
Therefore, according to the model, the population of the city in 2040 will be
54,775.
Question 27
Question
Solve the exponential equation for x:3x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: Since 27 can be written as 33, we have:
3x+1 = 33
Step 2: Now, we can set the exponents equal to each other:
x+ 1 = 3
Solving for x, we get
x= 3 −1
x= 2
Therefore, the solution to the equation is x= 2.
Question 28
Question
Solve the exponential equation 32x+1 = 27.
19
Solution
Step 1: Rewrite the equation in terms of a common base. Step 2: Solve for x.
Step 1: Rewrite the equation in terms of a common base. Since 27 = 33,
we can rewrite the equation 32x+1 = 27 as 32x+1 = 33.
Step 2: Solve for x. Since the bases are the same, we can set the exponents
equal to each other:
2x+ 1 = 3
Subtract 1 from both sides:
2x= 2
Divide by 2:
x= 1
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 29
Question
Solve the exponential equation 3x−1= 9.
Solution
Step 1: Rewrite 9as a power of 3.
3x−1= 32
Step 2: Since the bases are the same, set the exponents equal to each other.
(x−1) = 2
Step 3: Solve for x.
x= 2 + 1
Step 4: Simplify.
x= 3
Therefore, the solution to the exponential equation 3x−1= 9 is x= 3.
Question 30
Question
Solve the following exponential equation for x:
(5x) + (5x−1) = 150.
20
Solution
Step 1: Recall the properties of exponents to simplify the equation. Step 2:
Rewrite the equation in terms of a single base. Step 3: Solve the resulting
equation for x.
Step 1: Simplify the equation using the properties of exponents.
We know that am·an=am+n, so we can rewrite the given equation as:
5x+ 5x−1= 150.
Step 2: Rewrite the equation in terms of a single base.
We know that 5x−1=5x
5. Substituting this into the equation, we get:
5x+5x
5= 150.
Now, we can combine the terms:
5x+1
5·5x= 150.
Step 3: Solve the resulting equation for x.
Combining like terms, we have:
6
5·5x= 150.
Dividing by 6
5on both sides, we get:
5x=150 ·5
6= 125.
Taking the logarithm of both sides to solve for x, we have:
log5(5x) = log5(125).
This implies:
x= log5(125) = 3.
Therefore, the solution to the equation (5x) + (5x−1) = 150 is x= 3.
Question 31
Question
Solve the exponential equation 42x−1= 32.
21
Solution
Step 1: Rewrite 32 as a power of 4by noting that 32 = 42·2.
42x−1= 42·2
Step 2: Simplify the equation by equating the exponents.
2x−1 = 2 + 1
Step 3: Solve for xby isolating it.
2x= 3 + 1
2x= 4
Step 4: Divide by 2to solve for x.
x=4
2
Step 5: Simplify to find the final solution for x.
x= 2
Therefore, the solution to the exponential equation 42x−1= 32 is x= 2.
Question 32
Question
Solve the following exponential equation for x:3x+ 3x−1= 20.
Solution
Step 1: Rewrite 3x−1=1
3·3xto get the equation in terms of a single exponential
term. Step 2: Substitute 3x−1with 1
3·3xin the original equation to get 3x+
1
3·3x= 20. Step 3: Combine like terms to simplify the equation to 4
3·3x= 20.
Step 4: Multiply both sides by 3
4to isolate 3xon one side: 3x=15
2. Step 5:
Rewrite 15
2as 3log315
2. Step 6: Equate the exponents to solve for x:x= log315
2.
Step 7: Use a calculator to approximate x≈1.861.
Question 33
Question
Solve the exponential equation 3x−1+ 32x−1= 28.
22
Solution
Step 1: Let’s first rewrite 3x−1+ 32x−1= 28 in terms of the same base. We
can rewrite 32x−1as (32)x·3−1= 9x·1
3. Therefore, our equation becomes
3x−1+ 9x·1
3= 28.
Step 2: Multiply the entire equation by 3 to get rid of the fraction, resulting
in 3·3x−1+ 9x= 84.
Step 3: Utilize the properties of exponents to simplify the equation further.
3·3x−1= 31·3x−1= 3x. Therefore, the equation becomes 3x+ 9x= 84.
Step 4: Recognize that 9 = 32, so we can rewrite 9xas (32)x= 32x. Substi-
tuting this back into the equation gives us 3x+ 32x= 84.
Step 5: Factor out 3xfrom both terms to get 3x(1 + 3x) = 84.
Step 6: Now we can solve for 3xby dividing both sides by (1 + 3x), which
gives 3x=84
1+3x.
Step 7: Substitute y= 3x, which transforms the equation into y=84
1+y.
Step 8: Rearrange the equation to solve for y:y(1 + y) = 84. Simplify to
get y2+y−84 = 0.
Step 9: Factor the quadratic equation to get (y+ 12)(y−7) = 0. Therefore,
y=−12 or y= 7.
Step 10: Restore yto 3xto find that 3x=−12 does not have a real solution.
However, when 3x= 7, we find one valid solution: x= log37.
Question 34
Question
Solve the equation e2x= 10 for x. Round your answer to four decimal places if
necessary.
Solution
Step 1: Take the natural logarithm of both sides of the equation to eliminate
the base e.
ln(e2x)= ln(10)
Step 2: Use the property of logarithms that ln(ab)=bln(a).
2xln(e) = ln(10)
Step 3: Since ln(e) = 1, we have:
2x= ln(10)
Step 4: Divide by 2 to solve for x.
x=ln(10)
2≈0.6931
Therefore, the solution to the equation e2x= 10 is approximately x≈0.6931.
23
Question 35
Question
Samantha invested $10,000 in a savings account that earns 4.5% interest com-
pounded continuously. How long will it take for her investment to double in
value?
Solution
Let Arepresent the amount of money in the account at any time t, measured
in years. We know that the formula for continuous compound interest is given
by A=P ert, where: - Pis the initial principal amount ($10,000 in this case), -
ris the annual interest rate in decimal form (4.5% or 0.045 in this case), - tis
the time in years, and - eis the constant approximately equal to 2.71828.
We want to find the time it takes for the investment to double, so we are
looking for twhen A= 2P.
Step 1: Set up the equation for A:
A= 10000e0.045t
Step 2: Set Ato be double the initial amount:
2×10000 = 10000e0.045t
Step 3: Simplify the equation and solve for t:
20000 = 10000e0.045t
2 = e0.045t
Step 4: Take natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
ln(2) = 0.045t
t=ln(2)
0.045
Step 5: Calculate the value of t:
t≈ln(2)
0.045 ≈15.40 years
Therefore, it will take approximately 15.40 years for Samantha’s investment
to double in value when earning 4.5% interest compounded continuously.
24
Question 2
Question
Solve for x:23x+1 −5·2x+1 + 6 = 0.
Solution
Step 1: Let y= 2x. Then, rewrite the equation using y.
23x+1 −5·2x+1 + 6 = 0
⇒2·23x−5·2·2x+ 6 = 0
⇒2·(2x)3−5·2·2x+ 6 = 0
⇒2y3−10y+ 6 = 0
Step 2: Solve the cubic equation by factoring or using any appropriate
method.
Since the coefficient of y3is 2we can simplify the equation by dividing by
2:
y3−5y+ 3 = 0
Step 3: Try potential roots by using the Rational Root Theorem. The
possible rational roots are ±1,±3.
Substitute these values to find the root. We find y= 1 is a root by trial and
error.
Step 4: Use synthetic division to divide y3−5y+ 3 by (y−1).
1 1 0 −5
3
1 1
−4
Step 5: The synthetic division shows that y3−5y+ 3 = (y−1)(y2+y−3).
Step 6: Factor the quadratic polynomial y2+y−3.
y2+y−3 = (y+ 2)(y−1)
Step 7: Set each factor equal to zero and solve for y.
y+ 2 = 0 ⇒y=−2
y−1 = 0 ⇒y= 1
Step 8: Substitute back y= 2xto find the corresponding values of x:
y= 1 ⇒2x= 1 ⇒x= 0
y=−2⇒2x=−2(No real solution)
Step 9: Therefore, the solution to the equation 23x+1 −5·2x+1 + 6 = 0 is
x= 0.
2
Question 3
Question
Samantha invested $10,000 in a savings account that earns 4
Solution
Step 1: The formula for continuously compounded interest is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (initial investment). - ris the annual interest rate
(in decimal form). - tis the time the money is invested for in years. - eis Euler’s
number, approximately equal to 2.71828.
Step 2: Since Samantha wants to double her investment, the amount after t
years will be $20,000. So we have:
20,000 = 10,000 ·e0.04t
Step 3: Divide both sides by $10,000 to isolate the exponential term:
2 = e0.04t
Step 4: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.04t)
Step 5: Simplify using the properties of logarithms:
ln(2) = 0.04t·ln(e)
ln(2) = 0.04t
Step 6: Divide by 0.04 to solve for t:
t=ln(2)
0.04
Step 7: Calculate the approximate value of t:
t≈0.6931
0.04 ≈17.33
Step 8: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double in the savings account with 4
3
Question 4
Question
Samantha invested $10,000 in a savings account that pays an annual interest
rate of 5%, compounded continuously. How much money will be in the account
after 10 years?
Solution
Step 1: We can use the formula for compound interest compounded continu-
ously:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate (decimal), - tis the
time the money is invested for in years, and - eis Euler’s number, approximately
equal to 2.71828.
Step 2: Given that Samantha invests $10,000 at an annual interest rate of
5% (0.05 in decimal form), we have:
P= 10000, r = 0.05, t = 10
Plugging these values into the formula, we get:
A= 10000 ·e0.05·10
Step 3: Calculating the exponent, we have:
0.05 ·10 = 0.5
e0.5≈1.64872
Step 4: Finally, we can find the amount of money accumulated in the account
after 10 years:
A≈10000 ·1.64872 = 16487.20
Therefore, there will be approximately $16,487.20 in the account after 10
years.
Question 5
Question
A certain substance has a half-life of 3.5 hours. Suppose we start with an initial
amount of 100 grams of the substance. (a) Write an exponential decay model
for the amount of the substance remaining after thours. (b) How much of the
substance will remain after 10 hours? (c) How long will it take for the amount
of the substance to reduce to 25 grams?
4
Solution
(a) Let A(t)represent the amount of the substance remaining after thours. The
exponential decay model is given by:
A(t) = 100 ·(1
2)
t
3.5
(b) To find the remaining amount after 10 hours, we substitute t= 10 into
the model:
A(10) = 100 ·(1
2)10
3.5
A(10) = 100 ·(1
2)2.8571
A(10) ≈100 ·0.0786
A(10) ≈7.86 grams
(c) To find the time it takes for the amount to reduce to 25 grams, we set
A(t)to 25 and solve for t:
25 = 100 ·(1
2)
t
3.5
(1
2)
t
3.5
=1
4
t
3.5= 2
t= 7 ×3.5
t= 24.5hours
Question 6
Question
Solve the exponential equation: 3x−1= 27.
Solution
Step 1: Rewrite the equation in terms of a common base.
3x−1= 27
3x−1= 33
5
Step 2: Set the exponents equal to each other.
x−1 = 3
Step 3: Solve for x.
x= 3 + 1
x= 4
Step 4: Check the solution.
34−1= 27
33= 27
27 = 27
Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.
Question 7
Question
Let P(t) = 5e0.08trepresent the population of a town tyears after the year 2021.
Determine the year in which the population of the town is projected to reach
50,000.
Solution
Step 1: Set up the equation to find the year when the population is 50,000. Let
P(t) = 50,000 and solve for t.
5e0.08t= 50,000
Step 2: Divide both sides by 5 to isolate the exponential term.
e0.08t= 10,000
Step 3: Take the natural logarithm of both sides to solve for t.
ln(e0.08t)= ln(10,000)
0.08t= ln(10,000)
Step 4: Simplify the right side by using the property that ln(ab)=bln(a).
0.08t= 4 ln(10)
Step 5: Divide by 0.08 to solve for t.
t=4 ln(10)
0.08
6
Step 6: Calculate the value of t.
t=4×2.3026
0.08
t≈9.2104
0.08
t≈115.255
Step 7: Since trepresents the number of years from 2021, we add 2021 to
find the year when the population is projected to reach 50,000.
2021 + 115.255 ≈2136.255
Therefore, the population of the town is projected to reach 50,000 around
the year 2136.
Question 8
Question
Suppose a bacteria culture starts with 100 bacteria and triples every hour.
1. Find a formula for the number of bacteria after thours.
2. How many bacteria will there be after 5 hours?
Solution
1. Let P(t)represent the population of bacteria after thours. Since the
number of bacteria triples every hour, we have:
P(t) = 100 ×3t
Step 1: The formula for the number of bacteria after thours is P(t) =
100 ×3t.
2. To find the number of bacteria after 5 hours, we substitute t= 5 into the
formula:
P(5) = 100 ×35= 100 ×243 = 24,300
So, after 5 hours, there will be 24,300 bacteria.
Step 2: Substituting t= 5 into the formula, we find P(5) = 24,300.
Question 9
Question
The population of a town is currently 10,000 and is expected to triple every 20
years. Write an exponential function to model the population of the town after
tyears. What will be the population of the town in 60 years?
7
Solution
Step 1: Let P(t)be the population of the town after tyears. Since the population
is tripling every 20 years, the exponential model is given by P(t) = 10,000 ×
3t/20.
Step 2: To find the population of the town in 60 years, substitute t= 60
into the formula:
P(60) = 10,000 ×360/20
P(60) = 10,000 ×33
P(60) = 10,000 ×27
P(60) = 270,000
Therefore, the population of the town after 60 years will be 270,000.
Question 10
Question
Samantha deposits $5,000 into a savings account that pays 3.5% interest com-
pounded continuously. How much will be in the account after 10 years?
Solution
Step 1: We can use the formula for compound interest compounded continu-
ously:
A=P ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial deposit), r= annual interest rate (decimal), t= time in years,
and eis the base of the natural logarithm (approximately 2.71828).
Step 2: Given that: P= $5,000,r= 0.035 (3.5% expressed as a decimal),
and t= 10 years.
Step 3: Substituting the given values into the formula, we get:
A= 5000 ·e0.035·10
Step 4: Calculating the exponent:
A= 5000 ·e0.35
Step 5: Evaluating the exponential term:
A≈5000 ·1.419067
Step 6: Finally, we find the amount accumulated after 10 years:
A≈$7,095.34
Therefore, after 10 years, there will be approximately $7,095.34 in the ac-
count.
8
Question 11
Question
Solve for x:32x+1 = 54
Solution
Step 1: Rewrite 54 as a power of 3 by recognizing that 54 = 33.
32x+1 = 33
Step 2: Set the exponents equal, since the bases are the same.
2x+ 1 = 3
Step 3: Solve for x. Subtract 1 from both sides:
2x= 2
Divide by 2:
x= 1
Step 4: Therefore, the solution to the equation 32x+1 = 54 is x= 1 .
Question 12
Question
Solve the following logarithmic equation for x:3 log(x+ 2) −2 log(x) = 4.
Solution
Step 1: Use the properties of logarithms to simplify the equation. We know
that log(a)−log(b) = log (a
band log(an) = nlog(a). Applying these properties,
we get:
3 log(x+ 2) −2 log(x) = log((x+ 2)3)−log(x2)
Step 2: Simplify the equation further:
log((x+ 2)3)−log(x2)= 4
log ((x+ 2)3
x2)= 4
Step 3: Rewrite the equation in exponential form. We know that logb(a) = c
is equivalent to bc=a. Applying this to our equation, we have:
(x+ 2)3
x2= 104
9
Step 4: Simplify the right side:
(x+ 2)3= 104·x2
(x+ 2)3= 10000x2
Step 5: Expand the left side of the equation:
x3+ 6x2+ 12x+ 8 = 10000x2
Step 6: Rearrange the equation and set it equal to zero:
x3−9994x2+ 12x+ 8 = 0
Step 7: This equation doesn’t seem to simplify easily. To solve this equation,
we may need to use numerical methods or a calculator.
Therefore, the solution to the given logarithmic equation is xvalues that
satisfy the equation x3−9994x2+ 12x+ 8 = 0.
Question 13
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since the population was first measured.
Calculate the population of the city after 10 years.
Solution
Step 1: Substitute t= 10 into the population model:
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210:
1.0210 = 1.218391
Step 3: Substitute this value back into the population model to find P(10):
P(10) = 5000 ·1.218391 = 6091.955
Step 4: Therefore, after 10 years, the population of the city would be ap-
proximately 6092 people.
Question 14
Question
Samantha deposited $5000 into a savings account that earns 3.8% annual inter-
est, compounded quarterly. How much will be in the account after 10 years?
10
Solution
Step 1: Calculate the interest rate per compounding period. Given that the
annual interest rate is 3.8%, the quarterly interest rate is 3.8%/4=0.95% =
0.0095 in decimal form.
Step 2: Determine the number of compounding periods over 10 years. Since
the interest is compounded quarterly, there are 4compounding periods per year
and 4×10 = 40 total compounding periods over 10 years.
Step 3: Use the compound interest formula to find the future value of the
account. The formula for compound interest is:
A=P(1 + r
n)nt
where: - Ais the future value of the account, - Pis the principal amount ($5000
in this case), - ris the interest rate per compounding period (0.0095), - nis the
number of compounding periods per year (4), - and tis the number of years the
money is invested (10).
Substitute the values into the formula:
A= 5000 (1 + 0.0095
4)4×10
Step 4: Perform the calculations to find the future value of the account.
A= 5000 (1 + 0.002375)40
A= 5000 ×1.00237540
A≈5000 ×1.4141
A≈$7070.50
Therefore, after 10 years, there will be approximately $7070.50 in the ac-
count.
Question 15
Question
A certain population of bacteria triples every 5 hours. If there are initially 100
bacteria, how many bacteria will be present after 15 hours?
Solution
Step 1: Write down the formula for exponential growth. The formula for expo-
nential growth is given by:
N(t) = N0·at/d,
11
where: N(t)is the population at time t,N0is the initial population, ais the
growth factor, tis time elapsed, and dis the time it takes for the population to
grow by a factor of a.
Step 2: Find the growth factor. Since the population triples every 5 hours,
the growth factor ais 3, because 31= 3.
Step 3: Substitute the given values into the formula. Substitute N0= 100,
a= 3, and t= 15 into the formula to find N(15):
N(15) = 100 ·315
5.
Step 4: Calculate the population at 15 hours.
N(15) = 100 ·33= 100 ·27 = 2700.
Therefore, after 15 hours, there will be 2700 bacteria present.
Question 16
Question
Solve the following logarithmic equation for x:log2(3x+ 4) = log2(x+ 7).
Solution
Step 1: Since both sides of the equation are logarithms with base 2, we can
eliminate the logarithms by setting the arguments equal to each other:
3x+ 4 = x+ 7
Step 2: Next, we solve this linear equation for x:
3x+ 4 = x+ 7
2x= 3
x=3
2
Step 3: Therefore, the solution to the logarithmic equation log2(3x+ 4) =
log2(x+ 7) is x=3
2.
Question 17
Question
Suppose a researcher is studying the growth of a certain bacteria culture in a
controlled environment. The researcher determines that the bacteria population
doubles every 4 hours. If the initial population is 100 bacteria, what is the
exponential equation that models the population of the bacteria as a function
of time?
12
Solution
Step 1: Let’s denote the initial bacteria population as P0= 100 bacteria and
the time in hours as t. Since the population doubles every 4 hours, the growth
rate is 2.
Step 2: The exponential equation that models the population of the bacteria
as a function of time can be expressed as P(t) = P0·2t/4.
Step 3: Substitute P0= 100 into the equation to get P(t) = 100 ·2t/4.
Step 4: Therefore, the exponential equation that models the population of
the bacteria as a function of time is P(t) = 100 ·2t/4.
Question 18
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.03t, where
tis the number of years since the year 2020. Find the population of the city in
the year 2030.
Solution
Step 1: To find the population in the year 2030, we need to substitute t= 10
into the population function P(t). Step 2: Substitute t= 10 into the population
function:
P(10) = 5000 ·e0.03·10
Step 3: Simplify the expression:
P(10) = 5000 ·e0.3
Step 4: Use the fact that e≈2.71828 to approximate the population in the year
2030:
P(10) = 5000 ·2.718280.3
Step 5: Calculate the approximate population of the city in the year 2030:
P(10) ≈5000 ·1.349858
P(10) ≈6749.29
Step 6: Therefore, the population of the city in the year 2030 is approximately
6,749.29.
Question 19
Question
A certain radioactive substance decays according to the formula A(t) = A0e−kt,
where A(t)is the amount of substance remaining after tyears, A0is the initial
amount of substance, and kis a positive constant. If 80
13
Solution
Step 1: We are given that 80
0.20A0=A0e−10k
Dividing both sides by A0, we get:
0.20 = e−10k
Taking the natural logarithm of both sides, we have:
ln(0.20) = ln(e−10k)
ln(0.20) = −10k
k=−ln(0.20)
10
Step 2: Now, we want to find out how long it will take for 95
0.05A0=A0e−kt
0.05 = e−(−ln(0.20)
10 )t
0.05 = eln(0.20)
10 t
Step 3: To solve for t, take the natural logarithm of both sides:
ln(0.05) = ln(eln(0.20)
10 t)
ln(0.05) = ln(0.20)
10 t
t=10 ln(0.05)
ln(0.20)
Therefore, it will take approximately 10 ln(0.05)
ln(0.20) years for 95
Question 20
Question
The population of a city is modeled by the function P(t) = 5000 ·1.03t, where
P(t)represents the population after tyears. How long will it take for the
population to reach 10,000?
14
Solution
Step 1: Set up the equation by substituting P(t) = 10000 into the population
function:
10000 = 5000 ·1.03t
Step 2: Divide both sides by 5000 to isolate the exponential term:
10000
5000 = 1.03t
Step 3: Simplify the left side:
2 = 1.03t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(1.03t)
Step 5: Apply the properties of logarithms to bring down the exponent:
ln(2) = t·ln(1.03)
Step 6: Divide both sides by ln(1.03) to solve for t:
t=ln(2)
ln(1.03)
Step 7: Calculate the approximate value of tusing a calculator:
t≈ln(2)
ln(1.03) ≈22.44
So, it will take approximately 22.44 years for the population to reach 10,000
in the city.
Question 21
Question
Samantha has invested $5000 in a savings account that gives her an annual
interest rate of 5%, compounded continuously. How much will Samantha have
after 10 years?
Solution
Step 1: We can use the formula for continuous compound interest:
A=P·ert
15
where: - Ais the amount of money accumulated after tyears, - Pis the initial
investment ($5000 in this case), - ris the annual interest rate (expressed as a
decimal, so r= 0.05 in this case), and - tis the number of years the money is
invested for.
Step 2: Plug in the values into the formula to find the amount of money
Samantha will have after 10 years:
A= 5000 ·e0.05·10
Step 3: Calculate the value of the exponent:
0.05 ·10 = 0.5
Step 4: Substitute the exponent back into the formula:
A= 5000 ·e0.5
Step 5: Calculate the final amount by evaluating the exponential function:
A= 5000 ·e0.5≈5000 ·1.648721 ≈8243.61
Therefore, after 10 years, Samantha will have approximately $8243.61 in her
savings account.
Question 22
Question
The population of a city is growing exponentially with a growth rate of 3.5
Solution
Step 1: Determine the growth factor. Let P(t)represent the population at time
t, and rbe the annual growth rate in decimal form. The growth factor is given
by the formula:
F= 1 + r
In this case, r= 0.035, so the growth factor is F= 1 + 0.035 = 1.035.
Step 2: Find the population after 10 years. The population after tyears can
be modeled by the formula:
P(t) = P0·Ft
where P0is the initial population. Given that the current population is P0=
200,000 and we want to find the population in 10 years, we substitute t= 10
into the formula:
P(10) = 200,000 ·1.03510
Step 3: Calculate the population after 10 years. Now, we compute:
P(10) = 200,000 ·1.03510 ≈200,000 ·1.424 = 284,800
So, the estimated population of the city after 10 years will be approximately
284,800.
16
Question 23
Question
Solve the exponential equation 3x+1 = 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3. Since 27 = 33, the equation becomes
3x+1 = 33.
Step 2: Set the exponents equal to each other. We have x+ 1 = 3.
Step 3: Solve for xby subtracting 1 from both sides. We get x= 3 −1.
Step 4: Simplify to find the final answer. Therefore, x= 2.
Thus, the solution to the exponential equation 3x+1 = 27 is x= 2.
Question 24
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the exponents equal to each other.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
Step 4: Solve for x.
2x= 3 −1
2x= 2
x= 1
Step 5: Check the solution.
32(1)+1 = 33
33= 27
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
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Question 25
Question
Sara invested $5000 in a savings account that earns 2% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Let trepresent the time it takes for the investment to double in value. To solve
for t, we can use the continuous compound interest formula:
A=P·ert
Where: - Ais the amount of money after time t, - Pis the initial investment
amount, - ris the interest rate, - eis the base of the natural logarithm, - tis
the time.
In this case, Sara’s initial investment is $5000, the interest rate is 2% = 0.02,
and the amount she wants to reach is 2·$5000 = $10000.
Step 1: Set up the equation using the given information:
$10000 = $5000 ·e0.02t
Step 2: Divide both sides by $5000 to isolate e0.02t:
2 = e0.02t
Step 3: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.02t)
Step 4: Use the property of logarithms to simplify:
ln 2 = 0.02t·ln(e)
ln 2 = 0.02t
Step 5: Finally, solve for tby dividing both sides by 0.02:
t=ln 2
0.02
Thus, it will take approximately 34.66 years for Sara’s investment to double
in value.
Question 26
Question
The population of a city is modeled by the function P(t) = 25000 ·1.02t, where
trepresents the number of years since 2020. Calculate the population of the
city in 2040 according to this model.
18
Solution
Step 1: Determine the value of tin 2040. Since 2040 is 20 years after 2020,
t= 20.
Step 2: Substitute t= 20 into the population function.
P(20) = 25000 ·1.0220
Step 3: Calculate the population in 2040.
P(20) = 25000 ·1.0220 = 25000 ·2.191 = 54775
Therefore, according to the model, the population of the city in 2040 will be
54,775.
Question 27
Question
Solve the exponential equation for x:3x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: Since 27 can be written as 33, we have:
3x+1 = 33
Step 2: Now, we can set the exponents equal to each other:
x+ 1 = 3
Solving for x, we get
x= 3 −1
x= 2
Therefore, the solution to the equation is x= 2.
Question 28
Question
Solve the exponential equation 32x+1 = 27.
19
Solution
Step 1: Rewrite the equation in terms of a common base. Step 2: Solve for x.
Step 1: Rewrite the equation in terms of a common base. Since 27 = 33,
we can rewrite the equation 32x+1 = 27 as 32x+1 = 33.
Step 2: Solve for x. Since the bases are the same, we can set the exponents
equal to each other:
2x+ 1 = 3
Subtract 1 from both sides:
2x= 2
Divide by 2:
x= 1
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 29
Question
Solve the exponential equation 3x−1= 9.
Solution
Step 1: Rewrite 9as a power of 3.
3x−1= 32
Step 2: Since the bases are the same, set the exponents equal to each other.
(x−1) = 2
Step 3: Solve for x.
x= 2 + 1
Step 4: Simplify.
x= 3
Therefore, the solution to the exponential equation 3x−1= 9 is x= 3.
Question 30
Question
Solve the following exponential equation for x:
(5x) + (5x−1) = 150.
20
Solution
Step 1: Recall the properties of exponents to simplify the equation. Step 2:
Rewrite the equation in terms of a single base. Step 3: Solve the resulting
equation for x.
Step 1: Simplify the equation using the properties of exponents.
We know that am·an=am+n, so we can rewrite the given equation as:
5x+ 5x−1= 150.
Step 2: Rewrite the equation in terms of a single base.
We know that 5x−1=5x
5. Substituting this into the equation, we get:
5x+5x
5= 150.
Now, we can combine the terms:
5x+1
5·5x= 150.
Step 3: Solve the resulting equation for x.
Combining like terms, we have:
6
5·5x= 150.
Dividing by 6
5on both sides, we get:
5x=150 ·5
6= 125.
Taking the logarithm of both sides to solve for x, we have:
log5(5x) = log5(125).
This implies:
x= log5(125) = 3.
Therefore, the solution to the equation (5x) + (5x−1) = 150 is x= 3.
Question 31
Question
Solve the exponential equation 42x−1= 32.
21
Solution
Step 1: Rewrite 32 as a power of 4by noting that 32 = 42·2.
42x−1= 42·2
Step 2: Simplify the equation by equating the exponents.
2x−1 = 2 + 1
Step 3: Solve for xby isolating it.
2x= 3 + 1
2x= 4
Step 4: Divide by 2to solve for x.
x=4
2
Step 5: Simplify to find the final solution for x.
x= 2
Therefore, the solution to the exponential equation 42x−1= 32 is x= 2.
Question 32
Question
Solve the following exponential equation for x:3x+ 3x−1= 20.
Solution
Step 1: Rewrite 3x−1=1
3·3xto get the equation in terms of a single exponential
term. Step 2: Substitute 3x−1with 1
3·3xin the original equation to get 3x+
1
3·3x= 20. Step 3: Combine like terms to simplify the equation to 4
3·3x= 20.
Step 4: Multiply both sides by 3
4to isolate 3xon one side: 3x=15
2. Step 5:
Rewrite 15
2as 3log315
2. Step 6: Equate the exponents to solve for x:x= log315
2.
Step 7: Use a calculator to approximate x≈1.861.
Question 33
Question
Solve the exponential equation 3x−1+ 32x−1= 28.
22
Solution
Step 1: Let’s first rewrite 3x−1+ 32x−1= 28 in terms of the same base. We
can rewrite 32x−1as (32)x·3−1= 9x·1
3. Therefore, our equation becomes
3x−1+ 9x·1
3= 28.
Step 2: Multiply the entire equation by 3 to get rid of the fraction, resulting
in 3·3x−1+ 9x= 84.
Step 3: Utilize the properties of exponents to simplify the equation further.
3·3x−1= 31·3x−1= 3x. Therefore, the equation becomes 3x+ 9x= 84.
Step 4: Recognize that 9 = 32, so we can rewrite 9xas (32)x= 32x. Substi-
tuting this back into the equation gives us 3x+ 32x= 84.
Step 5: Factor out 3xfrom both terms to get 3x(1 + 3x) = 84.
Step 6: Now we can solve for 3xby dividing both sides by (1 + 3x), which
gives 3x=84
1+3x.
Step 7: Substitute y= 3x, which transforms the equation into y=84
1+y.
Step 8: Rearrange the equation to solve for y:y(1 + y) = 84. Simplify to
get y2+y−84 = 0.
Step 9: Factor the quadratic equation to get (y+ 12)(y−7) = 0. Therefore,
y=−12 or y= 7.
Step 10: Restore yto 3xto find that 3x=−12 does not have a real solution.
However, when 3x= 7, we find one valid solution: x= log37.
Question 34
Question
Solve the equation e2x= 10 for x. Round your answer to four decimal places if
necessary.
Solution
Step 1: Take the natural logarithm of both sides of the equation to eliminate
the base e.
ln(e2x)= ln(10)
Step 2: Use the property of logarithms that ln(ab)=bln(a).
2xln(e) = ln(10)
Step 3: Since ln(e) = 1, we have:
2x= ln(10)
Step 4: Divide by 2 to solve for x.
x=ln(10)
2≈0.6931
Therefore, the solution to the equation e2x= 10 is approximately x≈0.6931.
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Question 35
Question
Samantha invested $10,000 in a savings account that earns 4.5% interest com-
pounded continuously. How long will it take for her investment to double in
value?
Solution
Let Arepresent the amount of money in the account at any time t, measured
in years. We know that the formula for continuous compound interest is given
by A=P ert, where: - Pis the initial principal amount ($10,000 in this case), -
ris the annual interest rate in decimal form (4.5% or 0.045 in this case), - tis
the time in years, and - eis the constant approximately equal to 2.71828.
We want to find the time it takes for the investment to double, so we are
looking for twhen A= 2P.
Step 1: Set up the equation for A:
A= 10000e0.045t
Step 2: Set Ato be double the initial amount:
2×10000 = 10000e0.045t
Step 3: Simplify the equation and solve for t:
20000 = 10000e0.045t
2 = e0.045t
Step 4: Take natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
ln(2) = 0.045t
t=ln(2)
0.045
Step 5: Calculate the value of t:
t≈ln(2)
0.045 ≈15.40 years
Therefore, it will take approximately 15.40 years for Samantha’s investment
to double in value when earning 4.5% interest compounded continuously.
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