MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 8
Liberty University
Question 1
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4% compounded continuously. How much will the investment be worth after
10 years?
Solution
Step 1: We can use the formula for continuously compounded interest to find
the future value of the investment:
A=P·ert
where: Ais the amount of money accumulated after tyears, Pis the principal
amount (initial investment), ris the annual interest rate (in decimal form), and
tis the time the money is invested for.
Step 2: Substituting the given values into the formula, we have:
A= 5000 ·e0.04·10
Step 3: We can simplify this to find the final amount:
A= 5000 ·e0.4
Step 4: Calculate the value of the expression e0.4:
A≈5000 ·1.4918246948
Step 5: Multiply to find the final amount:
A≈7459.123474
Therefore, the investment will be worth approximately $7459.12 after 10
years.
Question 2
Question
The population of a city is modeled by the function P(t) = 2000 ·1.03t, where
trepresents the number of years since the initial population count.
a) What is the initial population of the city?
b) How long will it take for the population to triple in size?
Solution
a) Step 1: To find the initial population of the city, we can plug in t= 0 into
the given function P(t).
P(0) = 2000 ·1.030
= 2000 ·1
= 2000
Step 2: Therefore, the initial population of the city is 2000.
b) Step 1: To find out how long it will take for the population to triple in
size, we need to solve the equation P(t) = 3 ·P(0).
3·P(0) = 3 ·2000
= 6000
Step 2: Substitute P(t) = 2000 ·1.03tinto the equation and solve for t.
2000 ·1.03t= 6000
1.03t= 3
Step 3: Taking the natural logarithm of both sides gives:
ln(1.03t)= ln(3)
t·ln(1.03) = ln(3)
t=ln(3)
ln(1.03) ≈22.56
Step 4: Therefore, it will take approximately 22.56 years for the population
to triple in size.
Question 3
Question
Samantha invests $5000 in a savings account that earns 3% interest compounded
quarterly. How much money will she have in the account after 5 years?
2
Solution
Step 1: Identify the formula for compound interest: The formula for compound
interest is given by:
A=P(1 + r
n)nt
Where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate (in decimal form), -
nis the number of times interest is compounded per year, and - tis the number
of years the money is invested for.
Step 2: Substitute the given values into the formula: Given: - P= $5000 -
r= 0.03 (3% as a decimal) - n= 4 (compounded quarterly) - t= 5 years
Substitute these values into the formula to find the amount Samantha will
have after 5 years:
A= 5000 (1 + 0.03
4)4·5
Step 3: Calculate the amount in the account after 5 years: Calculating the
expression inside the parentheses first:
1 + 0.03
4= 1 + 0.0075 = 1.0075
Now, substitute this back into the formula:
A= 5000 ×(1.0075)20
Step 4: Compute the final amount: Calculating the final amount:
A≈5000 ×1.379008125 ≈$6895.04
Therefore, Samantha will have approximately $6895.04 in the account after
5 years.
Question 4
Question
Solve for x:52x+1 = 125
Solution
Step 1: Rewrite 125 as a power of 5. 125 = 53
Step 2: Set the two exponents equal to each other. 2x+ 1 = 3
Step 3: Solve for x.2x+ 1 = 3
2x= 2
x= 1
Therefore, the solution to the equation 52x+1 = 125 is x= 1.
3
Question 5
Question
Samantha invests $5,000 in a savings account that compounds continuously at
an annual interest rate of 4%. How long will it take for her investment to double
in value?
Solution
Step 1: Let trepresent the time (in years) it will take for Samantha’s investment
to double. We can use the continuous compounding formula:
A=P·ert,
where: - Ais the amount after tyears - Pis the principal amount (initial
investment) - ris the annual interest rate (expressed as a decimal) - tis the
time in years
Step 2: Since Samantha’s investment needs to double, the amount after t
years will be 2P. Substituting this information into the continuous compounding
formula, we get:
2P=P·e0.04t.
Step 3: Divide both sides of the equation by Pto simplify:
2 = e0.04t.
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.04t).
Step 5: Using the property of logarithms that ln(ex) = x, we simplify to:
ln(2) = 0.04t.
Step 6: Divide both sides by 0.04 to solve for t:
t=ln(2)
0.04 ≈0.6931
0.04 ≈17.33.
Step 7: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double in value.
Question 6
Question
Samantha invested $10,000 in an account that earns 7% interest compounded
continuously. How long will it take for Samantha’s investment to double in
value?
4
Solution
Step 1: The formula for compound interest with continuous compounding is
given by A=P ert, where: - Ais the amount of money accumulated after t
years, - 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 (≈2.71828).
Step 2: Since Samantha wants her investment to double, we set up the
equation 2P=P e0.07t.
Step 3: Simplifying the equation, we get 2 = e0.07t.
Step 4: Taking the natural logarithm of both sides, we have ln(2) = ln(e0.07t).
Step 5: By the properties of logarithms, ln(2) = 0.07tln(e).
Step 6: Since ln(e) = 1, we have ln(2) = 0.07t.
Step 7: Solving for t,t=ln(2)
0.07 .
Step 8: Using a calculator, t≈0.6931
0.07 ≈9.902.
Step 9: Therefore, it will take approximately 9.902 years for Samantha’s
investment to double in value.
Question 7
Question
Let f(x) = 3(2)x. Find the inverse function of f(x)and identify its domain and
range.
Solution
Step 1: To find the inverse function of f(x), we first replace f(x)with yand
then swap xand y.
y= 3(2)x
x= 3(2)y
Step 2: Next, we solve for y.
x
3= 2y
y= log2(x
3)
Therefore, the inverse function of f(x)is f−1(x) = log2(x
3).
Step 3: To identify the domain of f−1(x), we look at the restrictions on the
logarithmic function. Since the argument of a logarithm must be greater than
0, we have: x
3>0
x > 0
Thus, the domain of f−1(x)is all real numbers greater than 0, or x∈(0,∞).
5
Step 4: To determine the range of f−1(x), we look at the behavior of the
logarithmic function. As xapproaches 0, the logarithm approaches negative
infinity, and as xincreases towards infinity, the logarithm grows without bound.
Therefore, the range of f−1(x)is (−∞,∞).
Question 8
Question
Solve the exponential equation 22x+1 = 32.
Solution
Step 1: Rewrite 32 as a power of 2. We have 32 = 25.
Step 2: Substitute 32 with 25in the equation 22x+1 = 32. This gives us
22x+1 = 25.
Step 3: Since the bases are the same, set the exponents equal to each other:
2x+ 1 = 5.
Step 4: Solve for xby subtracting 1 from both sides: 2x= 4.
Step 5: Divide by 2 to solve for x:x= 2.
Therefore, the solution to the exponential equation 22x+1 = 32 is x= 2.
Question 9
Question
Solve the following exponential equation for x:23x+1 = 32.
Solution
Step 1: Rewrite 32 as a power of 2.
32 = 25
Step 2: Set the exponents equal to each other.
3x+ 1 = 5
Step 3: Solve for x.
3x= 4
x=4
3
Therefore, the solution to the exponential equation 23x+1 = 32 is x=4
3.
6
Question 10
Question
Solve the exponential equation 32x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation:
32x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
2x= 3
Step 4: Solve for xby dividing both sides by 2:
x=3
2
Therefore, the solution to the equation 32x= 27 is x=3
2.
Question 11
Question
Suppose a population of bacteria doubles every 2 hours. If there are initially 100
bacteria, how many bacteria will there be after 6 hours? Round your answer to
the nearest whole number.
Solution
Step 1: We can represent the population of bacteria after thours as an expo-
nential function. Let P(t)represent the population after thours. Since the
population doubles every 2 hours, we have that P(t) = 100 ×2t/2.
Step 2: To find the population of bacteria after 6 hours, we need to evaluate
P(6). Substituting t= 6 into the function, we get P(6) = 100×26/2= 100×23=
100 ×8 = 800.
Therefore, after 6 hours, there will be approximately 800 bacteria.
7
Question 12
Question
Samantha invests $5000 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, which is given by
A=P ert,
where: - Ais the amount of money accumulated after tyears, - Pis the princi-
pal amount (initial investment), - ris the annual interest rate (expressed as a
decimal), - tis the time the money is invested for, in years, and - eis Euler’s
number (≈2.71828).
Step 2: Substitute P= 5000,r= 0.05, and t= 10 into the formula.
A= 5000 ·e0.05·10
Step 3: Calculate the amount of money in the account after 10 years.
A= 5000 ·e0.5
Step 4: Use a calculator to find the approximate value of e0.5.
A≈5000 ·1.64872 ≈8243.6
Therefore, after 10 years, there will be approximately $8243.60 in the ac-
count.
Question 13
Question
Let f(x) = 3(2)xand g(x) = log2(x−1). Find the values of xfor which
f(x) = g(x).
Solution
Step 1: Set f(x) = g(x)and substitute the expressions for f(x)and g(x)into
the equation:
(3)(2)x= log2(x−1)
Step 2: Since the base of the logarithm is 2, rewrite the exponential expres-
sion using base 2:
3(2)x= log2(x−1) =⇒3(2)x=ln(x−1)
ln(2)
8
Step 3: Now, solve the equation by changing the exponential expression to
a logarithmic expression:
ln(3(2)x) = ln(ln(x−1)
ln(2) )
Step 4: Use the properties of logarithms to simplify the equation:
ln(3) + ln(2x) = ln(ln(x−1)) −ln(ln(2))
ln(3) + xln(2) = ln(ln(x−1)) −ln(ln(2))
Step 5: Further simplify the equation to isolate x:
xln(2) = ln(ln(x−1)) −ln(ln(2)) −ln(3)
x=ln(ln(x−1)) −ln(ln(2)) −ln(3)
ln(2)
Therefore, the solutions for xare the values that satisfy this final equation.
Question 14
Question
The population of a city is modeled by the function P(t) = 2000 ·1.03t, where
P(t)represents the population after tyears. If the current population of the
city is 2300, find after how many years the population is expected to reach 3000.
Solution
Step 1: First, we substitute the given value of the current population into the
function P(t)and solve for t:
2300 = 2000 ·1.03t
Step 2: Divide both sides by 2000 to get:
1.15 = 1.03t
Step 3: Take the natural logarithm of both sides to solve for t:
ln(1.15) = ln(1.03t)
Step 4: Use the property ln(ab)=b·ln(a)and simplify the right side:
ln(1.15) = t·ln(1.03)
Step 5: Solve for tby dividing both sides by ln(1.03):
t=ln(1.15)
ln(1.03)
Step 6: Using a calculator, we find t≈7.52 years.
Therefore, the population of the city is expected to reach 3000 after approx-
imately 7.52 years.
9
Question 15
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite both sides of the equation with the same base. 27 can be
written as 33, so the equation becomes 32x−1= 33.
Step 2: Since the bases are the same, the exponents must be equal. Set
2x−1 = 3.
Step 3: Solve for xby isolating x. Add 1to both sides: 2x= 4.
Step 4: Divide by 2to solve for x:x= 2.
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 16
Question
The population of a certain city is modeled by the function P(t) = 5000e0.02t,
where trepresents the number of years since the initial population count was
taken.
How long will it take for the population to reach 10,000?
Solution
Step 1: Set up the equation P(t) = 10000 and solve for t:
5000e0.02t= 10000
Step 2: Divide both sides by 5000 to isolate the exponential term:
e0.02t= 2
Step 3: Take the natural logarithm of both sides to eliminate the exponential
term:
ln(e0.02t)= ln(2)
Step 4: Use the property of logarithms that ln(ex) = x:
0.02t= ln(2)
Step 5: Solve for tby dividing both sides by 0.02:
t=ln(2)
0.02
10
Step 6: Use a calculator to find the approximate value:
t≈ln(2)
0.02 ≈0.6931
0.02 ≈34.655
So, it will take approximately 34.655 years for the population to reach 10,000.
Question 17
Question
Solve the exponential equation 2x−1= 8.
Solution
Step 1: Rewrite 8 as a power of 2.
2x−1= 23
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 by substituting x= 4 back into the original
equation.
24−1= 23
23= 8
Therefore, the solution to the exponential equation is x= 4.
Question 18
Question
Samantha invested $10,000 in an account that earns an annual interest rate
of 5% compounded continuously. How long will it take for her investment to
double?
11
Solution
Let P(t)be the amount of money in the account at time t(in years) after the
initial investment. The formula for continuous compounding is given by:
P(t) = P0·ert
where: - P0is the initial investment, - ris the annual interest rate, decimal
form, - tis time in years, - eis the base of the natural logarithm, approximately
equal to 2.71828.
In this case, we are given that P0= $10,000,r= 0.05, and we want to find
tsuch that P(t) = 2 ·P0= 2 ·10,000 = 20,000.
Step 1: Plug in the given values and the desired future amount into the
formula.
20,000 = 10,000 ·e0.05t
Step 2: Divide both sides by 10,000 to isolate the exponential term.
2 = e0.05t
Step 3: Take the natural logarithm of both sides to solve for t.
ln(2) = ln(e0.05t)
Step 4: Remember that ln(ex) = x, so we can simplify to:
ln(2) = 0.05t
Step 5: Now, solve for t:
t=ln(2)
0.05 ≈0.6931
0.05 ≈13.86 years
Therefore, it will take approximately 13.86 years for Samantha’s investment
to double.
Question 19
Question
Samantha invests $500 in a savings account that earns interest compounded
continuously. If the account is worth $650.60 after 3 years, what is the annual
interest rate? Round your answer to two decimal places.
Solution
Step 1: The formula for continuously compounded interest is given by the for-
mula A=P ert, where: - Ais the amount of money accumulated after time t,
12
-Pis the principal amount (initial investment), - ris the annual interest rate,
and - tis the time the money is invested for.
Step 2: We are given that: - P= $500, - A= $650.60, and - t= 3 years.
Step 3: Substituting these values into the formula, we get:
650.60 = 500e3r
Step 4: Divide both sides by 500:
650.60
500 =e3r
Step 5: Simplify the left side:
1.3012 = e3r
Step 6: Take the natural logarithm of both sides to solve for r:
ln(1.3012) = ln(e3r)
Step 7: Use the property ln(ex) = x:
ln(1.3012) = 3r
Step 8: Solve for r:
r=ln(1.3012)
3≈0.0977
Step 9: Therefore, the annual interest rate is approximately 9.77%.
Question 20
Question
Solve the following exponential equation for x:
32x−1−3x−1−20 = 0
Solution
Step 1: Let’s first rewrite the equation using exponent properties:
32x−1−3x−1−20 = 0
32x·3−1−3x·3−1−20 = 0
32x·1
3−3x·1
3−20 = 0
Step 2: Simplify the equation:
32x
3−3x
3−20 = 0
13
32x−1−3x−1−20 = 0
Step 3: Let u= 3x, then the equation becomes:
u2−u−20 = 0
Step 4: Factor the quadratic equation:
(u−5)(u+ 4) = 0
Step 5: Solve for u:
u−5 = 0 or u+ 4 = 0
u= 5 or u=−4
Step 6: Substitute back to find x: For u= 5:
3x= 5
x= log35
For u=−4, this solution is not valid since u= 3xmust be positive.
Therefore, the solution to the equation is x= log35.
Question 21
Question
Solve the exponential equation 2x+1 −2x= 24.
Solution
Step 1: Rewrite the equation using the properties of exponents. Step 2: Solve
for 2xby combining like terms. Step 3: Solve for xusing logarithms.
Step 1: Rewrite the equation using the properties of exponents. We can
rewrite 2x+1 as 2x·2. So the equation becomes 2x·2−2x= 24.
Step 2: Solve for 2xby combining like terms. Subtracting 2xfrom both
sides of the equation gives us 2·2x−2x= 24 which simplifies to 2x= 24.
Step 3: Solve for xusing logarithms. Taking the logarithm of both sides,
we get log(2x) = log(24). Using the property of logarithm log(ab)=blog(a),
we can rewrite this as xlog(2) = log(24). Finally, solving for xgives x=log(24)
log(2) .
Therefore, the solution to the exponential equation 2x+1 −2x= 24 is x=
log(24)
log(2) .
Question 22
Question
Solve the exponential equation 22x+1 = 16 for x.
14
Solution
Step 1: Rewrite 16 as a power of 2.
16 = 24
Step 2: Set up the equation with the bases equal to each other.
22x+1 = 24
Step 3: Since the bases are equal, set the exponents equal to each other.
2x+ 1 = 4
Step 4: Solve for x.
2x= 4 −1
2x= 3
x=3
2
Step 5: Therefore, the solution to the equation 22x+1 = 16 is x=3
2.
Question 23
Question
Assume that the population of a city follows an exponential growth model, where
the population doubles every 15 years. If the current population is 200,000, write
the exponential growth formula for the population of the city.
Solution
Step 1: Let P(t)be the population of the city at time t. Since the population
doubles every 15 years, the growth factor is 2. Therefore, the exponential growth
formula can be written as:
P(t) = 200,000 ×2t/15
Step 2: Simplify the formula by substituting P(t) = 200,000 and t= 0 to
find the initial population:
200,000 = 200,000 ×20/15
200,000 = 200,000 ×20
200,000 = 200,000
Step 3: Therefore, the exponential growth formula for the population of the
city is P(t) = 200,000 ×2t/15.
15
Question 24
Question
Solve the exponential equation 2x−3= 5.
Solution
Step 1: Take the logarithm of both sides to eliminate the exponent. Choose a
logarithm base that will help simplify the equation.
log2(2x−3) = log2(5)
Step 2: Apply the power rule of logarithms to bring down the exponent.
(x−3) ·log2(2) = log2(5)
Step 3: Since log2(2) = 1, simplify the equation.
x−3 = log2(5)
Step 4: Add 3 to both sides to isolate x.
x= log2(5) + 3
Step 5: Use the change of base formula for logarithms to rewrite the equation
in a common base (e.g., base 10 or base e).
x=ln(5)
ln(2) + 3
Therefore, the solution to the exponential equation 2x−3= 5 is x=ln(5)
ln(2) +3.
Question 25
Question
Solve the exponential equation 1
2·3x+1 = 27.
Solution
Step 1: Start by expressing 27 as a power of 3.
1
2·3x+1 = 27
3x+1 = 2 ·27
3x+1 = 54
16
Step 2: Rewrite 54 as a power of 3.
3x+1 = 33·2
3x+1 = 33·31
3x+1 = 33+1
Step 3: Equate the exponents and solve for x.
x+ 1 = 3 + 1
x+ 1 = 4
x= 4 −1
x= 3
Step 4: Check the solution by substituting x= 3 back into the original
equation. 1
2·33+1 = 27
1
2·34= 27
1
2·81 = 27
40.5 = 27
Step 5: Since the last step yielded a false statement, there is no solution to
the equation 1
2·3x+1 = 27.
Question 26
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
We know that 27 = 33. So, we can rewrite the equation as 32x−1= 33.
Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal to each other:
2x−1 = 3.
Step 3: Solve for x.
Add 1to both sides of the equation to isolate 2x:2x= 4.
Divide by 2to solve for x:x= 2.
Step 4: Check the solution.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
Simplify both sides: 33= 33.
Since the left side is equal to the right side, the solution x= 2 is verified.
17
Question 27
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation 32x−1= 27:
32x−1= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
2x−1 = 3
Step 4: Solve for xby adding 1to both sides and then dividing by 2:
2x−1 = 3
2x= 4
x= 2
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 28
Question
Solve for x:2x+3 −4·2x+1 + 4 = 0.
Solution
Step 1: Let’s rewrite the equation in terms of 2xto make the manipulation
easier. Let y= 2x. Then the equation becomes:
y2−4y+ 4 = 0
Step 2: Now we need to solve the quadratic equation y2−4y+ 4 = 0.
Step 3: Factoring the quadratic, we get (y−2)2= 0.
Step 4: Taking the square root of both sides, we find y−2 = 0, so y= 2.
Step 5: Recall that y= 2x, so 2x= 2.
Step 6: Solving for x, we find x= 1.
Therefore, the solution to the equation 2x+3 −4·2x+1 + 4 = 0 is x= 1.
18
Question 29
Question
Solve the following exponential equation for x:3x−1−3x= 3.
Solution
Step 1: Let’s start by rewriting the equation using a common base. We know
that 3x−1=3x
3. Substituting this into the equation, we get 3x
3−3x= 3.
Step 2: Next, let’s combine the fractions on the left side of the equation. We
have 3x
−3·3x
3= 3.
Step 3: Simplifying further gives us −2·3x
3= 3.
Step 4: Multiplying both sides by 3, we get −2·3x= 9.
Step 5: Now, divide by -2 to solve for 3x. We have 3x=−9
2.
Step 6: Since 3xshould be a positive real number, there is no real solution
for xin this case. Therefore, the equation 3x−1−3x= 3 has no real solution.
Question 30
Question
Samantha invests $5000 in an account that earns 3.5% interest compounded
continuously. How long will it take for Samantha’s investment to double?
Solution
Step 1: Determine the formula to calculate the amount in the account after
time tyears with continuous compounding. The formula is given by:
A=P·ert
where: A= amount in the account after tyears, P= initial investment, r
= interest rate per year, t= time in years, and eis the base of the natural
logarithm, approximately equal to 2.71828.
Step 2: For this scenario, Samantha wants her investment to double, so the
amount in the account (A) should be equal to twice the initial investment (2P):
2P=P·e0.035t
Step 3: Divide both sides of the equation by Pto simplify:
2 = e0.035t
Step 4: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.035t)
19
ln(2) = 0.035tln(e)
ln(2) = 0.035t
Step 5: Divide by 0.035 to solve for t:
t=ln(2)
0.035 ≈19.81 years
Therefore, it will take approximately 19.81 years for Samantha’s investment
to double in the account with continuous compounding.
Question 31
Question
A bacteria culture initially contains 100 bacteria and doubles in size every 2
hours.
(a) Find a formula for the number of bacteria after t hours.
(b) How many bacteria will be present after 6 hours?
Solution
(a) Let N(t)represent the number of bacteria after thours. Since the bacteria
culture doubles in size every 2 hours, we have:
N(t) = 100 ·2t
2
(b) To find the number of bacteria after 6 hours, we substitute t= 6 into
the formula we found in part (a):
N(6) = 100 ·26
2
N(6) = 100 ·23
N(6) = 100 ·8
N(6) = 800
After 6 hours, there will be 800 bacteria present in the culture.
Question 32
Question
Samantha invests $5000 in a savings account that earns 4% interest compounded
continuously. How long will it take for her money to double?
20
Solution
Step 1: The formula for continuously compounded interest is given by A=P ert,
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate in decimal form, - t
is the time the money is invested for, - eis Euler’s number (≈2.71828).
Step 2: In this case, Samantha invested P= $5000, the interest rate is
r= 0.04, and she wants her money to double. Therefore, the amount she wants
is 2P= $10000.
Step 3: Plugging in the given values into the continuously compounded
interest formula, we have:
10000 = 5000 ·e0.04t
Step 4: Divide both sides by 5000:
2 = e0.04t
Step 5: To solve for t, we take the natural logarithm of both sides:
ln(2) = ln(e0.04t)
Step 6: Remember that ln(ex) = x, so we get:
ln(2) = 0.04t
Step 7: Now, isolate tby dividing by 0.04:
t=ln(2)
0.04
Step 8: Using a calculator, we find:
t≈0.693147
0.04 ≈17.33
Step 9: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double.
Question 33
Question
Solve the exponential equation 32x+1 = 27.
21
Solution
Step 1: Rewrite 27 as a power of 3:27 = 33.
Step 2: Substitute 33into the original equation: 32x+1 = 33.
Step 3: Since the bases are the same, we can set the exponents equal to each
other: 2x+ 1 = 3.
Step 4: Solve for xby isolating the variable:
2x+ 1 = 3
2x= 3 −1
2x= 2
x=2
2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation:
32(1)+1 = 27
33= 27
27 = 27
Since the solution satisfies the original equation, x= 1 is the correct solution.
Question 34
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: We can rewrite 27 as 33. Therefore, the equation becomes:
3x−1= 33
Step 2: Since the base is the same, we can set the exponents equal to each
other:
x−1 = 3
Now, solve for x:
x= 3 + 1
x= 4
Therefore, the solution to the equation 3x−1= 27 is x= 4.
22
Question 35
Question
A certain investment grows according to the formula A(t) = 5000 ·e0.04t, where
A(t)represents the amount of money in the account after tyears. Calculate the
time it will take for the investment to double.
Solution
Step 1: To find when the investment doubles, we need to set A(t)equal to twice
the initial amount:
2·5000 = 5000 ·e0.04t
Step 2: Simplify the equation:
10000 = 5000 ·e0.04t
Step 3: Divide both sides by 5000 to isolate the exponential term:
2 = e0.04t
Step 4: To solve for t, take the natural logarithm of both sides to undo the
exponential:
ln(2) = ln(e0.04t)
Step 5: By the property of logarithms, we can bring down the exponent:
ln(2) = 0.04tln(e)
Step 6: Since ln(e) = 1, the equation becomes:
ln(2) = 0.04t
Step 7: Finally, divide by 0.04 to solve for t:
t=ln(2)
0.04 ≈0.6931
0.04 ≈17.33 years
Therefore, it will take approximately 17.33 years for the investment to dou-
ble.
23
Question 2
Question
The population of a city is modeled by the function P(t) = 2000 ·1.03t, where
trepresents the number of years since the initial population count.
a) What is the initial population of the city?
b) How long will it take for the population to triple in size?
Solution
a) Step 1: To find the initial population of the city, we can plug in t= 0 into
the given function P(t).
P(0) = 2000 ·1.030
= 2000 ·1
= 2000
Step 2: Therefore, the initial population of the city is 2000.
b) Step 1: To find out how long it will take for the population to triple in
size, we need to solve the equation P(t) = 3 ·P(0).
3·P(0) = 3 ·2000
= 6000
Step 2: Substitute P(t) = 2000 ·1.03tinto the equation and solve for t.
2000 ·1.03t= 6000
1.03t= 3
Step 3: Taking the natural logarithm of both sides gives:
ln(1.03t)= ln(3)
t·ln(1.03) = ln(3)
t=ln(3)
ln(1.03) ≈22.56
Step 4: Therefore, it will take approximately 22.56 years for the population
to triple in size.
Question 3
Question
Samantha invests $5000 in a savings account that earns 3% interest compounded
quarterly. How much money will she have in the account after 5 years?
2
Solution
Step 1: Identify the formula for compound interest: The formula for compound
interest is given by:
A=P(1 + r
n)nt
Where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate (in decimal form), -
nis the number of times interest is compounded per year, and - tis the number
of years the money is invested for.
Step 2: Substitute the given values into the formula: Given: - P= $5000 -
r= 0.03 (3% as a decimal) - n= 4 (compounded quarterly) - t= 5 years
Substitute these values into the formula to find the amount Samantha will
have after 5 years:
A= 5000 (1 + 0.03
4)4·5
Step 3: Calculate the amount in the account after 5 years: Calculating the
expression inside the parentheses first:
1 + 0.03
4= 1 + 0.0075 = 1.0075
Now, substitute this back into the formula:
A= 5000 ×(1.0075)20
Step 4: Compute the final amount: Calculating the final amount:
A≈5000 ×1.379008125 ≈$6895.04
Therefore, Samantha will have approximately $6895.04 in the account after
5 years.
Question 4
Question
Solve for x:52x+1 = 125
Solution
Step 1: Rewrite 125 as a power of 5. 125 = 53
Step 2: Set the two exponents equal to each other. 2x+ 1 = 3
Step 3: Solve for x.2x+ 1 = 3
2x= 2
x= 1
Therefore, the solution to the equation 52x+1 = 125 is x= 1.
3
Question 5
Question
Samantha invests $5,000 in a savings account that compounds continuously at
an annual interest rate of 4%. How long will it take for her investment to double
in value?
Solution
Step 1: Let trepresent the time (in years) it will take for Samantha’s investment
to double. We can use the continuous compounding formula:
A=P·ert,
where: - Ais the amount after tyears - Pis the principal amount (initial
investment) - ris the annual interest rate (expressed as a decimal) - tis the
time in years
Step 2: Since Samantha’s investment needs to double, the amount after t
years will be 2P. Substituting this information into the continuous compounding
formula, we get:
2P=P·e0.04t.
Step 3: Divide both sides of the equation by Pto simplify:
2 = e0.04t.
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.04t).
Step 5: Using the property of logarithms that ln(ex) = x, we simplify to:
ln(2) = 0.04t.
Step 6: Divide both sides by 0.04 to solve for t:
t=ln(2)
0.04 ≈0.6931
0.04 ≈17.33.
Step 7: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double in value.
Question 6
Question
Samantha invested $10,000 in an account that earns 7% interest compounded
continuously. How long will it take for Samantha’s investment to double in
value?
4
Solution
Step 1: The formula for compound interest with continuous compounding is
given by A=P ert, where: - Ais the amount of money accumulated after t
years, - 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 (≈2.71828).
Step 2: Since Samantha wants her investment to double, we set up the
equation 2P=P e0.07t.
Step 3: Simplifying the equation, we get 2 = e0.07t.
Step 4: Taking the natural logarithm of both sides, we have ln(2) = ln(e0.07t).
Step 5: By the properties of logarithms, ln(2) = 0.07tln(e).
Step 6: Since ln(e) = 1, we have ln(2) = 0.07t.
Step 7: Solving for t,t=ln(2)
0.07 .
Step 8: Using a calculator, t≈0.6931
0.07 ≈9.902.
Step 9: Therefore, it will take approximately 9.902 years for Samantha’s
investment to double in value.
Question 7
Question
Let f(x) = 3(2)x. Find the inverse function of f(x)and identify its domain and
range.
Solution
Step 1: To find the inverse function of f(x), we first replace f(x)with yand
then swap xand y.
y= 3(2)x
x= 3(2)y
Step 2: Next, we solve for y.
x
3= 2y
y= log2(x
3)
Therefore, the inverse function of f(x)is f−1(x) = log2(x
3).
Step 3: To identify the domain of f−1(x), we look at the restrictions on the
logarithmic function. Since the argument of a logarithm must be greater than
0, we have: x
3>0
x > 0
Thus, the domain of f−1(x)is all real numbers greater than 0, or x∈(0,∞).
5
Step 4: To determine the range of f−1(x), we look at the behavior of the
logarithmic function. As xapproaches 0, the logarithm approaches negative
infinity, and as xincreases towards infinity, the logarithm grows without bound.
Therefore, the range of f−1(x)is (−∞,∞).
Question 8
Question
Solve the exponential equation 22x+1 = 32.
Solution
Step 1: Rewrite 32 as a power of 2. We have 32 = 25.
Step 2: Substitute 32 with 25in the equation 22x+1 = 32. This gives us
22x+1 = 25.
Step 3: Since the bases are the same, set the exponents equal to each other:
2x+ 1 = 5.
Step 4: Solve for xby subtracting 1 from both sides: 2x= 4.
Step 5: Divide by 2 to solve for x:x= 2.
Therefore, the solution to the exponential equation 22x+1 = 32 is x= 2.
Question 9
Question
Solve the following exponential equation for x:23x+1 = 32.
Solution
Step 1: Rewrite 32 as a power of 2.
32 = 25
Step 2: Set the exponents equal to each other.
3x+ 1 = 5
Step 3: Solve for x.
3x= 4
x=4
3
Therefore, the solution to the exponential equation 23x+1 = 32 is x=4
3.
6
Question 10
Question
Solve the exponential equation 32x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation:
32x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
2x= 3
Step 4: Solve for xby dividing both sides by 2:
x=3
2
Therefore, the solution to the equation 32x= 27 is x=3
2.
Question 11
Question
Suppose a population of bacteria doubles every 2 hours. If there are initially 100
bacteria, how many bacteria will there be after 6 hours? Round your answer to
the nearest whole number.
Solution
Step 1: We can represent the population of bacteria after thours as an expo-
nential function. Let P(t)represent the population after thours. Since the
population doubles every 2 hours, we have that P(t) = 100 ×2t/2.
Step 2: To find the population of bacteria after 6 hours, we need to evaluate
P(6). Substituting t= 6 into the function, we get P(6) = 100×26/2= 100×23=
100 ×8 = 800.
Therefore, after 6 hours, there will be approximately 800 bacteria.
7
Question 12
Question
Samantha invests $5000 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, which is given by
A=P ert,
where: - Ais the amount of money accumulated after tyears, - Pis the princi-
pal amount (initial investment), - ris the annual interest rate (expressed as a
decimal), - tis the time the money is invested for, in years, and - eis Euler’s
number (≈2.71828).
Step 2: Substitute P= 5000,r= 0.05, and t= 10 into the formula.
A= 5000 ·e0.05·10
Step 3: Calculate the amount of money in the account after 10 years.
A= 5000 ·e0.5
Step 4: Use a calculator to find the approximate value of e0.5.
A≈5000 ·1.64872 ≈8243.6
Therefore, after 10 years, there will be approximately $8243.60 in the ac-
count.
Question 13
Question
Let f(x) = 3(2)xand g(x) = log2(x−1). Find the values of xfor which
f(x) = g(x).
Solution
Step 1: Set f(x) = g(x)and substitute the expressions for f(x)and g(x)into
the equation:
(3)(2)x= log2(x−1)
Step 2: Since the base of the logarithm is 2, rewrite the exponential expres-
sion using base 2:
3(2)x= log2(x−1) =⇒3(2)x=ln(x−1)
ln(2)
8
Step 3: Now, solve the equation by changing the exponential expression to
a logarithmic expression:
ln(3(2)x) = ln(ln(x−1)
ln(2) )
Step 4: Use the properties of logarithms to simplify the equation:
ln(3) + ln(2x) = ln(ln(x−1)) −ln(ln(2))
ln(3) + xln(2) = ln(ln(x−1)) −ln(ln(2))
Step 5: Further simplify the equation to isolate x:
xln(2) = ln(ln(x−1)) −ln(ln(2)) −ln(3)
x=ln(ln(x−1)) −ln(ln(2)) −ln(3)
ln(2)
Therefore, the solutions for xare the values that satisfy this final equation.
Question 14
Question
The population of a city is modeled by the function P(t) = 2000 ·1.03t, where
P(t)represents the population after tyears. If the current population of the
city is 2300, find after how many years the population is expected to reach 3000.
Solution
Step 1: First, we substitute the given value of the current population into the
function P(t)and solve for t:
2300 = 2000 ·1.03t
Step 2: Divide both sides by 2000 to get:
1.15 = 1.03t
Step 3: Take the natural logarithm of both sides to solve for t:
ln(1.15) = ln(1.03t)
Step 4: Use the property ln(ab)=b·ln(a)and simplify the right side:
ln(1.15) = t·ln(1.03)
Step 5: Solve for tby dividing both sides by ln(1.03):
t=ln(1.15)
ln(1.03)
Step 6: Using a calculator, we find t≈7.52 years.
Therefore, the population of the city is expected to reach 3000 after approx-
imately 7.52 years.
9
Question 15
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite both sides of the equation with the same base. 27 can be
written as 33, so the equation becomes 32x−1= 33.
Step 2: Since the bases are the same, the exponents must be equal. Set
2x−1 = 3.
Step 3: Solve for xby isolating x. Add 1to both sides: 2x= 4.
Step 4: Divide by 2to solve for x:x= 2.
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 16
Question
The population of a certain city is modeled by the function P(t) = 5000e0.02t,
where trepresents the number of years since the initial population count was
taken.
How long will it take for the population to reach 10,000?
Solution
Step 1: Set up the equation P(t) = 10000 and solve for t:
5000e0.02t= 10000
Step 2: Divide both sides by 5000 to isolate the exponential term:
e0.02t= 2
Step 3: Take the natural logarithm of both sides to eliminate the exponential
term:
ln(e0.02t)= ln(2)
Step 4: Use the property of logarithms that ln(ex) = x:
0.02t= ln(2)
Step 5: Solve for tby dividing both sides by 0.02:
t=ln(2)
0.02
10
Step 6: Use a calculator to find the approximate value:
t≈ln(2)
0.02 ≈0.6931
0.02 ≈34.655
So, it will take approximately 34.655 years for the population to reach 10,000.
Question 17
Question
Solve the exponential equation 2x−1= 8.
Solution
Step 1: Rewrite 8 as a power of 2.
2x−1= 23
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 by substituting x= 4 back into the original
equation.
24−1= 23
23= 8
Therefore, the solution to the exponential equation is x= 4.
Question 18
Question
Samantha invested $10,000 in an account that earns an annual interest rate
of 5% compounded continuously. How long will it take for her investment to
double?
11
Solution
Let P(t)be the amount of money in the account at time t(in years) after the
initial investment. The formula for continuous compounding is given by:
P(t) = P0·ert
where: - P0is the initial investment, - ris the annual interest rate, decimal
form, - tis time in years, - eis the base of the natural logarithm, approximately
equal to 2.71828.
In this case, we are given that P0= $10,000,r= 0.05, and we want to find
tsuch that P(t) = 2 ·P0= 2 ·10,000 = 20,000.
Step 1: Plug in the given values and the desired future amount into the
formula.
20,000 = 10,000 ·e0.05t
Step 2: Divide both sides by 10,000 to isolate the exponential term.
2 = e0.05t
Step 3: Take the natural logarithm of both sides to solve for t.
ln(2) = ln(e0.05t)
Step 4: Remember that ln(ex) = x, so we can simplify to:
ln(2) = 0.05t
Step 5: Now, solve for t:
t=ln(2)
0.05 ≈0.6931
0.05 ≈13.86 years
Therefore, it will take approximately 13.86 years for Samantha’s investment
to double.
Question 19
Question
Samantha invests $500 in a savings account that earns interest compounded
continuously. If the account is worth $650.60 after 3 years, what is the annual
interest rate? Round your answer to two decimal places.
Solution
Step 1: The formula for continuously compounded interest is given by the for-
mula A=P ert, where: - Ais the amount of money accumulated after time t,
12
-Pis the principal amount (initial investment), - ris the annual interest rate,
and - tis the time the money is invested for.
Step 2: We are given that: - P= $500, - A= $650.60, and - t= 3 years.
Step 3: Substituting these values into the formula, we get:
650.60 = 500e3r
Step 4: Divide both sides by 500:
650.60
500 =e3r
Step 5: Simplify the left side:
1.3012 = e3r
Step 6: Take the natural logarithm of both sides to solve for r:
ln(1.3012) = ln(e3r)
Step 7: Use the property ln(ex) = x:
ln(1.3012) = 3r
Step 8: Solve for r:
r=ln(1.3012)
3≈0.0977
Step 9: Therefore, the annual interest rate is approximately 9.77%.
Question 20
Question
Solve the following exponential equation for x:
32x−1−3x−1−20 = 0
Solution
Step 1: Let’s first rewrite the equation using exponent properties:
32x−1−3x−1−20 = 0
32x·3−1−3x·3−1−20 = 0
32x·1
3−3x·1
3−20 = 0
Step 2: Simplify the equation:
32x
3−3x
3−20 = 0
13
32x−1−3x−1−20 = 0
Step 3: Let u= 3x, then the equation becomes:
u2−u−20 = 0
Step 4: Factor the quadratic equation:
(u−5)(u+ 4) = 0
Step 5: Solve for u:
u−5 = 0 or u+ 4 = 0
u= 5 or u=−4
Step 6: Substitute back to find x: For u= 5:
3x= 5
x= log35
For u=−4, this solution is not valid since u= 3xmust be positive.
Therefore, the solution to the equation is x= log35.
Question 21
Question
Solve the exponential equation 2x+1 −2x= 24.
Solution
Step 1: Rewrite the equation using the properties of exponents. Step 2: Solve
for 2xby combining like terms. Step 3: Solve for xusing logarithms.
Step 1: Rewrite the equation using the properties of exponents. We can
rewrite 2x+1 as 2x·2. So the equation becomes 2x·2−2x= 24.
Step 2: Solve for 2xby combining like terms. Subtracting 2xfrom both
sides of the equation gives us 2·2x−2x= 24 which simplifies to 2x= 24.
Step 3: Solve for xusing logarithms. Taking the logarithm of both sides,
we get log(2x) = log(24). Using the property of logarithm log(ab)=blog(a),
we can rewrite this as xlog(2) = log(24). Finally, solving for xgives x=log(24)
log(2) .
Therefore, the solution to the exponential equation 2x+1 −2x= 24 is x=
log(24)
log(2) .
Question 22
Question
Solve the exponential equation 22x+1 = 16 for x.
14
Solution
Step 1: Rewrite 16 as a power of 2.
16 = 24
Step 2: Set up the equation with the bases equal to each other.
22x+1 = 24
Step 3: Since the bases are equal, set the exponents equal to each other.
2x+ 1 = 4
Step 4: Solve for x.
2x= 4 −1
2x= 3
x=3
2
Step 5: Therefore, the solution to the equation 22x+1 = 16 is x=3
2.
Question 23
Question
Assume that the population of a city follows an exponential growth model, where
the population doubles every 15 years. If the current population is 200,000, write
the exponential growth formula for the population of the city.
Solution
Step 1: Let P(t)be the population of the city at time t. Since the population
doubles every 15 years, the growth factor is 2. Therefore, the exponential growth
formula can be written as:
P(t) = 200,000 ×2t/15
Step 2: Simplify the formula by substituting P(t) = 200,000 and t= 0 to
find the initial population:
200,000 = 200,000 ×20/15
200,000 = 200,000 ×20
200,000 = 200,000
Step 3: Therefore, the exponential growth formula for the population of the
city is P(t) = 200,000 ×2t/15.
15
Question 24
Question
Solve the exponential equation 2x−3= 5.
Solution
Step 1: Take the logarithm of both sides to eliminate the exponent. Choose a
logarithm base that will help simplify the equation.
log2(2x−3) = log2(5)
Step 2: Apply the power rule of logarithms to bring down the exponent.
(x−3) ·log2(2) = log2(5)
Step 3: Since log2(2) = 1, simplify the equation.
x−3 = log2(5)
Step 4: Add 3 to both sides to isolate x.
x= log2(5) + 3
Step 5: Use the change of base formula for logarithms to rewrite the equation
in a common base (e.g., base 10 or base e).
x=ln(5)
ln(2) + 3
Therefore, the solution to the exponential equation 2x−3= 5 is x=ln(5)
ln(2) +3.
Question 25
Question
Solve the exponential equation 1
2·3x+1 = 27.
Solution
Step 1: Start by expressing 27 as a power of 3.
1
2·3x+1 = 27
3x+1 = 2 ·27
3x+1 = 54
16
Step 2: Rewrite 54 as a power of 3.
3x+1 = 33·2
3x+1 = 33·31
3x+1 = 33+1
Step 3: Equate the exponents and solve for x.
x+ 1 = 3 + 1
x+ 1 = 4
x= 4 −1
x= 3
Step 4: Check the solution by substituting x= 3 back into the original
equation. 1
2·33+1 = 27
1
2·34= 27
1
2·81 = 27
40.5 = 27
Step 5: Since the last step yielded a false statement, there is no solution to
the equation 1
2·3x+1 = 27.
Question 26
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
We know that 27 = 33. So, we can rewrite the equation as 32x−1= 33.
Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal to each other:
2x−1 = 3.
Step 3: Solve for x.
Add 1to both sides of the equation to isolate 2x:2x= 4.
Divide by 2to solve for x:x= 2.
Step 4: Check the solution.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
Simplify both sides: 33= 33.
Since the left side is equal to the right side, the solution x= 2 is verified.
17
Question 27
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation 32x−1= 27:
32x−1= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
2x−1 = 3
Step 4: Solve for xby adding 1to both sides and then dividing by 2:
2x−1 = 3
2x= 4
x= 2
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 28
Question
Solve for x:2x+3 −4·2x+1 + 4 = 0.
Solution
Step 1: Let’s rewrite the equation in terms of 2xto make the manipulation
easier. Let y= 2x. Then the equation becomes:
y2−4y+ 4 = 0
Step 2: Now we need to solve the quadratic equation y2−4y+ 4 = 0.
Step 3: Factoring the quadratic, we get (y−2)2= 0.
Step 4: Taking the square root of both sides, we find y−2 = 0, so y= 2.
Step 5: Recall that y= 2x, so 2x= 2.
Step 6: Solving for x, we find x= 1.
Therefore, the solution to the equation 2x+3 −4·2x+1 + 4 = 0 is x= 1.
18
Question 29
Question
Solve the following exponential equation for x:3x−1−3x= 3.
Solution
Step 1: Let’s start by rewriting the equation using a common base. We know
that 3x−1=3x
3. Substituting this into the equation, we get 3x
3−3x= 3.
Step 2: Next, let’s combine the fractions on the left side of the equation. We
have 3x
−3·3x
3= 3.
Step 3: Simplifying further gives us −2·3x
3= 3.
Step 4: Multiplying both sides by 3, we get −2·3x= 9.
Step 5: Now, divide by -2 to solve for 3x. We have 3x=−9
2.
Step 6: Since 3xshould be a positive real number, there is no real solution
for xin this case. Therefore, the equation 3x−1−3x= 3 has no real solution.
Question 30
Question
Samantha invests $5000 in an account that earns 3.5% interest compounded
continuously. How long will it take for Samantha’s investment to double?
Solution
Step 1: Determine the formula to calculate the amount in the account after
time tyears with continuous compounding. The formula is given by:
A=P·ert
where: A= amount in the account after tyears, P= initial investment, r
= interest rate per year, t= time in years, and eis the base of the natural
logarithm, approximately equal to 2.71828.
Step 2: For this scenario, Samantha wants her investment to double, so the
amount in the account (A) should be equal to twice the initial investment (2P):
2P=P·e0.035t
Step 3: Divide both sides of the equation by Pto simplify:
2 = e0.035t
Step 4: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.035t)
19
ln(2) = 0.035tln(e)
ln(2) = 0.035t
Step 5: Divide by 0.035 to solve for t:
t=ln(2)
0.035 ≈19.81 years
Therefore, it will take approximately 19.81 years for Samantha’s investment
to double in the account with continuous compounding.
Question 31
Question
A bacteria culture initially contains 100 bacteria and doubles in size every 2
hours.
(a) Find a formula for the number of bacteria after t hours.
(b) How many bacteria will be present after 6 hours?
Solution
(a) Let N(t)represent the number of bacteria after thours. Since the bacteria
culture doubles in size every 2 hours, we have:
N(t) = 100 ·2t
2
(b) To find the number of bacteria after 6 hours, we substitute t= 6 into
the formula we found in part (a):
N(6) = 100 ·26
2
N(6) = 100 ·23
N(6) = 100 ·8
N(6) = 800
After 6 hours, there will be 800 bacteria present in the culture.
Question 32
Question
Samantha invests $5000 in a savings account that earns 4% interest compounded
continuously. How long will it take for her money to double?
20
Solution
Step 1: The formula for continuously compounded interest is given by A=P ert,
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate in decimal form, - t
is the time the money is invested for, - eis Euler’s number (≈2.71828).
Step 2: In this case, Samantha invested P= $5000, the interest rate is
r= 0.04, and she wants her money to double. Therefore, the amount she wants
is 2P= $10000.
Step 3: Plugging in the given values into the continuously compounded
interest formula, we have:
10000 = 5000 ·e0.04t
Step 4: Divide both sides by 5000:
2 = e0.04t
Step 5: To solve for t, we take the natural logarithm of both sides:
ln(2) = ln(e0.04t)
Step 6: Remember that ln(ex) = x, so we get:
ln(2) = 0.04t
Step 7: Now, isolate tby dividing by 0.04:
t=ln(2)
0.04
Step 8: Using a calculator, we find:
t≈0.693147
0.04 ≈17.33
Step 9: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double.
Question 33
Question
Solve the exponential equation 32x+1 = 27.
21
Solution
Step 1: Rewrite 27 as a power of 3:27 = 33.
Step 2: Substitute 33into the original equation: 32x+1 = 33.
Step 3: Since the bases are the same, we can set the exponents equal to each
other: 2x+ 1 = 3.
Step 4: Solve for xby isolating the variable:
2x+ 1 = 3
2x= 3 −1
2x= 2
x=2
2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation:
32(1)+1 = 27
33= 27
27 = 27
Since the solution satisfies the original equation, x= 1 is the correct solution.
Question 34
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: We can rewrite 27 as 33. Therefore, the equation becomes:
3x−1= 33
Step 2: Since the base is the same, we can set the exponents equal to each
other:
x−1 = 3
Now, solve for x:
x= 3 + 1
x= 4
Therefore, the solution to the equation 3x−1= 27 is x= 4.
22
Question 35
Question
A certain investment grows according to the formula A(t) = 5000 ·e0.04t, where
A(t)represents the amount of money in the account after tyears. Calculate the
time it will take for the investment to double.
Solution
Step 1: To find when the investment doubles, we need to set A(t)equal to twice
the initial amount:
2·5000 = 5000 ·e0.04t
Step 2: Simplify the equation:
10000 = 5000 ·e0.04t
Step 3: Divide both sides by 5000 to isolate the exponential term:
2 = e0.04t
Step 4: To solve for t, take the natural logarithm of both sides to undo the
exponential:
ln(2) = ln(e0.04t)
Step 5: By the property of logarithms, we can bring down the exponent:
ln(2) = 0.04tln(e)
Step 6: Since ln(e) = 1, the equation becomes:
ln(2) = 0.04t
Step 7: Finally, divide by 0.04 to solve for t:
t=ln(2)
0.04 ≈0.6931
0.04 ≈17.33 years
Therefore, it will take approximately 17.33 years for the investment to dou-
ble.
23
Question 2
Question
The population of a city is modeled by the function P(t) = 2000 ·1.03t, where
trepresents the number of years since the initial population count.
a) What is the initial population of the city?
b) How long will it take for the population to triple in size?
Solution
a) Step 1: To find the initial population of the city, we can plug in t= 0 into
the given function P(t).
P(0) = 2000 ·1.030
= 2000 ·1
= 2000
Step 2: Therefore, the initial population of the city is 2000.
b) Step 1: To find out how long it will take for the population to triple in
size, we need to solve the equation P(t) = 3 ·P(0).
3·P(0) = 3 ·2000
= 6000
Step 2: Substitute P(t) = 2000 ·1.03tinto the equation and solve for t.
2000 ·1.03t= 6000
1.03t= 3
Step 3: Taking the natural logarithm of both sides gives:
ln(1.03t)= ln(3)
t·ln(1.03) = ln(3)
t=ln(3)
ln(1.03) ≈22.56
Step 4: Therefore, it will take approximately 22.56 years for the population
to triple in size.
Question 3
Question
Samantha invests $5000 in a savings account that earns 3% interest compounded
quarterly. How much money will she have in the account after 5 years?
2
Solution
Step 1: Identify the formula for compound interest: The formula for compound
interest is given by:
A=P(1 + r
n)nt
Where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate (in decimal form), -
nis the number of times interest is compounded per year, and - tis the number
of years the money is invested for.
Step 2: Substitute the given values into the formula: Given: - P= $5000 -
r= 0.03 (3% as a decimal) - n= 4 (compounded quarterly) - t= 5 years
Substitute these values into the formula to find the amount Samantha will
have after 5 years:
A= 5000 (1 + 0.03
4)4·5
Step 3: Calculate the amount in the account after 5 years: Calculating the
expression inside the parentheses first:
1 + 0.03
4= 1 + 0.0075 = 1.0075
Now, substitute this back into the formula:
A= 5000 ×(1.0075)20
Step 4: Compute the final amount: Calculating the final amount:
A≈5000 ×1.379008125 ≈$6895.04
Therefore, Samantha will have approximately $6895.04 in the account after
5 years.
Question 4
Question
Solve for x:52x+1 = 125
Solution
Step 1: Rewrite 125 as a power of 5. 125 = 53
Step 2: Set the two exponents equal to each other. 2x+ 1 = 3
Step 3: Solve for x.2x+ 1 = 3
2x= 2
x= 1
Therefore, the solution to the equation 52x+1 = 125 is x= 1.
3
Question 5
Question
Samantha invests $5,000 in a savings account that compounds continuously at
an annual interest rate of 4%. How long will it take for her investment to double
in value?
Solution
Step 1: Let trepresent the time (in years) it will take for Samantha’s investment
to double. We can use the continuous compounding formula:
A=P·ert,
where: - Ais the amount after tyears - Pis the principal amount (initial
investment) - ris the annual interest rate (expressed as a decimal) - tis the
time in years
Step 2: Since Samantha’s investment needs to double, the amount after t
years will be 2P. Substituting this information into the continuous compounding
formula, we get:
2P=P·e0.04t.
Step 3: Divide both sides of the equation by Pto simplify:
2 = e0.04t.
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.04t).
Step 5: Using the property of logarithms that ln(ex) = x, we simplify to:
ln(2) = 0.04t.
Step 6: Divide both sides by 0.04 to solve for t:
t=ln(2)
0.04 ≈0.6931
0.04 ≈17.33.
Step 7: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double in value.
Question 6
Question
Samantha invested $10,000 in an account that earns 7% interest compounded
continuously. How long will it take for Samantha’s investment to double in
value?
4
Solution
Step 1: The formula for compound interest with continuous compounding is
given by A=P ert, where: - Ais the amount of money accumulated after t
years, - 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 (≈2.71828).
Step 2: Since Samantha wants her investment to double, we set up the
equation 2P=P e0.07t.
Step 3: Simplifying the equation, we get 2 = e0.07t.
Step 4: Taking the natural logarithm of both sides, we have ln(2) = ln(e0.07t).
Step 5: By the properties of logarithms, ln(2) = 0.07tln(e).
Step 6: Since ln(e) = 1, we have ln(2) = 0.07t.
Step 7: Solving for t,t=ln(2)
0.07 .
Step 8: Using a calculator, t≈0.6931
0.07 ≈9.902.
Step 9: Therefore, it will take approximately 9.902 years for Samantha’s
investment to double in value.
Question 7
Question
Let f(x) = 3(2)x. Find the inverse function of f(x)and identify its domain and
range.
Solution
Step 1: To find the inverse function of f(x), we first replace f(x)with yand
then swap xand y.
y= 3(2)x
x= 3(2)y
Step 2: Next, we solve for y.
x
3= 2y
y= log2(x
3)
Therefore, the inverse function of f(x)is f−1(x) = log2(x
3).
Step 3: To identify the domain of f−1(x), we look at the restrictions on the
logarithmic function. Since the argument of a logarithm must be greater than
0, we have: x
3>0
x > 0
Thus, the domain of f−1(x)is all real numbers greater than 0, or x∈(0,∞).
5
Step 4: To determine the range of f−1(x), we look at the behavior of the
logarithmic function. As xapproaches 0, the logarithm approaches negative
infinity, and as xincreases towards infinity, the logarithm grows without bound.
Therefore, the range of f−1(x)is (−∞,∞).
Question 8
Question
Solve the exponential equation 22x+1 = 32.
Solution
Step 1: Rewrite 32 as a power of 2. We have 32 = 25.
Step 2: Substitute 32 with 25in the equation 22x+1 = 32. This gives us
22x+1 = 25.
Step 3: Since the bases are the same, set the exponents equal to each other:
2x+ 1 = 5.
Step 4: Solve for xby subtracting 1 from both sides: 2x= 4.
Step 5: Divide by 2 to solve for x:x= 2.
Therefore, the solution to the exponential equation 22x+1 = 32 is x= 2.
Question 9
Question
Solve the following exponential equation for x:23x+1 = 32.
Solution
Step 1: Rewrite 32 as a power of 2.
32 = 25
Step 2: Set the exponents equal to each other.
3x+ 1 = 5
Step 3: Solve for x.
3x= 4
x=4
3
Therefore, the solution to the exponential equation 23x+1 = 32 is x=4
3.
6
Question 10
Question
Solve the exponential equation 32x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation:
32x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
2x= 3
Step 4: Solve for xby dividing both sides by 2:
x=3
2
Therefore, the solution to the equation 32x= 27 is x=3
2.
Question 11
Question
Suppose a population of bacteria doubles every 2 hours. If there are initially 100
bacteria, how many bacteria will there be after 6 hours? Round your answer to
the nearest whole number.
Solution
Step 1: We can represent the population of bacteria after thours as an expo-
nential function. Let P(t)represent the population after thours. Since the
population doubles every 2 hours, we have that P(t) = 100 ×2t/2.
Step 2: To find the population of bacteria after 6 hours, we need to evaluate
P(6). Substituting t= 6 into the function, we get P(6) = 100×26/2= 100×23=
100 ×8 = 800.
Therefore, after 6 hours, there will be approximately 800 bacteria.
7
Question 12
Question
Samantha invests $5000 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, which is given by
A=P ert,
where: - Ais the amount of money accumulated after tyears, - Pis the princi-
pal amount (initial investment), - ris the annual interest rate (expressed as a
decimal), - tis the time the money is invested for, in years, and - eis Euler’s
number (≈2.71828).
Step 2: Substitute P= 5000,r= 0.05, and t= 10 into the formula.
A= 5000 ·e0.05·10
Step 3: Calculate the amount of money in the account after 10 years.
A= 5000 ·e0.5
Step 4: Use a calculator to find the approximate value of e0.5.
A≈5000 ·1.64872 ≈8243.6
Therefore, after 10 years, there will be approximately $8243.60 in the ac-
count.
Question 13
Question
Let f(x) = 3(2)xand g(x) = log2(x−1). Find the values of xfor which
f(x) = g(x).
Solution
Step 1: Set f(x) = g(x)and substitute the expressions for f(x)and g(x)into
the equation:
(3)(2)x= log2(x−1)
Step 2: Since the base of the logarithm is 2, rewrite the exponential expres-
sion using base 2:
3(2)x= log2(x−1) =⇒3(2)x=ln(x−1)
ln(2)
8
Step 3: Now, solve the equation by changing the exponential expression to
a logarithmic expression:
ln(3(2)x) = ln(ln(x−1)
ln(2) )
Step 4: Use the properties of logarithms to simplify the equation:
ln(3) + ln(2x) = ln(ln(x−1)) −ln(ln(2))
ln(3) + xln(2) = ln(ln(x−1)) −ln(ln(2))
Step 5: Further simplify the equation to isolate x:
xln(2) = ln(ln(x−1)) −ln(ln(2)) −ln(3)
x=ln(ln(x−1)) −ln(ln(2)) −ln(3)
ln(2)
Therefore, the solutions for xare the values that satisfy this final equation.
Question 14
Question
The population of a city is modeled by the function P(t) = 2000 ·1.03t, where
P(t)represents the population after tyears. If the current population of the
city is 2300, find after how many years the population is expected to reach 3000.
Solution
Step 1: First, we substitute the given value of the current population into the
function P(t)and solve for t:
2300 = 2000 ·1.03t
Step 2: Divide both sides by 2000 to get:
1.15 = 1.03t
Step 3: Take the natural logarithm of both sides to solve for t:
ln(1.15) = ln(1.03t)
Step 4: Use the property ln(ab)=b·ln(a)and simplify the right side:
ln(1.15) = t·ln(1.03)
Step 5: Solve for tby dividing both sides by ln(1.03):
t=ln(1.15)
ln(1.03)
Step 6: Using a calculator, we find t≈7.52 years.
Therefore, the population of the city is expected to reach 3000 after approx-
imately 7.52 years.
9
Question 15
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite both sides of the equation with the same base. 27 can be
written as 33, so the equation becomes 32x−1= 33.
Step 2: Since the bases are the same, the exponents must be equal. Set
2x−1 = 3.
Step 3: Solve for xby isolating x. Add 1to both sides: 2x= 4.
Step 4: Divide by 2to solve for x:x= 2.
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 16
Question
The population of a certain city is modeled by the function P(t) = 5000e0.02t,
where trepresents the number of years since the initial population count was
taken.
How long will it take for the population to reach 10,000?
Solution
Step 1: Set up the equation P(t) = 10000 and solve for t:
5000e0.02t= 10000
Step 2: Divide both sides by 5000 to isolate the exponential term:
e0.02t= 2
Step 3: Take the natural logarithm of both sides to eliminate the exponential
term:
ln(e0.02t)= ln(2)
Step 4: Use the property of logarithms that ln(ex) = x:
0.02t= ln(2)
Step 5: Solve for tby dividing both sides by 0.02:
t=ln(2)
0.02
10
Step 6: Use a calculator to find the approximate value:
t≈ln(2)
0.02 ≈0.6931
0.02 ≈34.655
So, it will take approximately 34.655 years for the population to reach 10,000.
Question 17
Question
Solve the exponential equation 2x−1= 8.
Solution
Step 1: Rewrite 8 as a power of 2.
2x−1= 23
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 by substituting x= 4 back into the original
equation.
24−1= 23
23= 8
Therefore, the solution to the exponential equation is x= 4.
Question 18
Question
Samantha invested $10,000 in an account that earns an annual interest rate
of 5% compounded continuously. How long will it take for her investment to
double?
11
Solution
Let P(t)be the amount of money in the account at time t(in years) after the
initial investment. The formula for continuous compounding is given by:
P(t) = P0·ert
where: - P0is the initial investment, - ris the annual interest rate, decimal
form, - tis time in years, - eis the base of the natural logarithm, approximately
equal to 2.71828.
In this case, we are given that P0= $10,000,r= 0.05, and we want to find
tsuch that P(t) = 2 ·P0= 2 ·10,000 = 20,000.
Step 1: Plug in the given values and the desired future amount into the
formula.
20,000 = 10,000 ·e0.05t
Step 2: Divide both sides by 10,000 to isolate the exponential term.
2 = e0.05t
Step 3: Take the natural logarithm of both sides to solve for t.
ln(2) = ln(e0.05t)
Step 4: Remember that ln(ex) = x, so we can simplify to:
ln(2) = 0.05t
Step 5: Now, solve for t:
t=ln(2)
0.05 ≈0.6931
0.05 ≈13.86 years
Therefore, it will take approximately 13.86 years for Samantha’s investment
to double.
Question 19
Question
Samantha invests $500 in a savings account that earns interest compounded
continuously. If the account is worth $650.60 after 3 years, what is the annual
interest rate? Round your answer to two decimal places.
Solution
Step 1: The formula for continuously compounded interest is given by the for-
mula A=P ert, where: - Ais the amount of money accumulated after time t,
12
-Pis the principal amount (initial investment), - ris the annual interest rate,
and - tis the time the money is invested for.
Step 2: We are given that: - P= $500, - A= $650.60, and - t= 3 years.
Step 3: Substituting these values into the formula, we get:
650.60 = 500e3r
Step 4: Divide both sides by 500:
650.60
500 =e3r
Step 5: Simplify the left side:
1.3012 = e3r
Step 6: Take the natural logarithm of both sides to solve for r:
ln(1.3012) = ln(e3r)
Step 7: Use the property ln(ex) = x:
ln(1.3012) = 3r
Step 8: Solve for r:
r=ln(1.3012)
3≈0.0977
Step 9: Therefore, the annual interest rate is approximately 9.77%.
Question 20
Question
Solve the following exponential equation for x:
32x−1−3x−1−20 = 0
Solution
Step 1: Let’s first rewrite the equation using exponent properties:
32x−1−3x−1−20 = 0
32x·3−1−3x·3−1−20 = 0
32x·1
3−3x·1
3−20 = 0
Step 2: Simplify the equation:
32x
3−3x
3−20 = 0
13
32x−1−3x−1−20 = 0
Step 3: Let u= 3x, then the equation becomes:
u2−u−20 = 0
Step 4: Factor the quadratic equation:
(u−5)(u+ 4) = 0
Step 5: Solve for u:
u−5 = 0 or u+ 4 = 0
u= 5 or u=−4
Step 6: Substitute back to find x: For u= 5:
3x= 5
x= log35
For u=−4, this solution is not valid since u= 3xmust be positive.
Therefore, the solution to the equation is x= log35.
Question 21
Question
Solve the exponential equation 2x+1 −2x= 24.
Solution
Step 1: Rewrite the equation using the properties of exponents. Step 2: Solve
for 2xby combining like terms. Step 3: Solve for xusing logarithms.
Step 1: Rewrite the equation using the properties of exponents. We can
rewrite 2x+1 as 2x·2. So the equation becomes 2x·2−2x= 24.
Step 2: Solve for 2xby combining like terms. Subtracting 2xfrom both
sides of the equation gives us 2·2x−2x= 24 which simplifies to 2x= 24.
Step 3: Solve for xusing logarithms. Taking the logarithm of both sides,
we get log(2x) = log(24). Using the property of logarithm log(ab)=blog(a),
we can rewrite this as xlog(2) = log(24). Finally, solving for xgives x=log(24)
log(2) .
Therefore, the solution to the exponential equation 2x+1 −2x= 24 is x=
log(24)
log(2) .
Question 22
Question
Solve the exponential equation 22x+1 = 16 for x.
14
Solution
Step 1: Rewrite 16 as a power of 2.
16 = 24
Step 2: Set up the equation with the bases equal to each other.
22x+1 = 24
Step 3: Since the bases are equal, set the exponents equal to each other.
2x+ 1 = 4
Step 4: Solve for x.
2x= 4 −1
2x= 3
x=3
2
Step 5: Therefore, the solution to the equation 22x+1 = 16 is x=3
2.
Question 23
Question
Assume that the population of a city follows an exponential growth model, where
the population doubles every 15 years. If the current population is 200,000, write
the exponential growth formula for the population of the city.
Solution
Step 1: Let P(t)be the population of the city at time t. Since the population
doubles every 15 years, the growth factor is 2. Therefore, the exponential growth
formula can be written as:
P(t) = 200,000 ×2t/15
Step 2: Simplify the formula by substituting P(t) = 200,000 and t= 0 to
find the initial population:
200,000 = 200,000 ×20/15
200,000 = 200,000 ×20
200,000 = 200,000
Step 3: Therefore, the exponential growth formula for the population of the
city is P(t) = 200,000 ×2t/15.
15
Question 24
Question
Solve the exponential equation 2x−3= 5.
Solution
Step 1: Take the logarithm of both sides to eliminate the exponent. Choose a
logarithm base that will help simplify the equation.
log2(2x−3) = log2(5)
Step 2: Apply the power rule of logarithms to bring down the exponent.
(x−3) ·log2(2) = log2(5)
Step 3: Since log2(2) = 1, simplify the equation.
x−3 = log2(5)
Step 4: Add 3 to both sides to isolate x.
x= log2(5) + 3
Step 5: Use the change of base formula for logarithms to rewrite the equation
in a common base (e.g., base 10 or base e).
x=ln(5)
ln(2) + 3
Therefore, the solution to the exponential equation 2x−3= 5 is x=ln(5)
ln(2) +3.
Question 25
Question
Solve the exponential equation 1
2·3x+1 = 27.
Solution
Step 1: Start by expressing 27 as a power of 3.
1
2·3x+1 = 27
3x+1 = 2 ·27
3x+1 = 54
16
Step 2: Rewrite 54 as a power of 3.
3x+1 = 33·2
3x+1 = 33·31
3x+1 = 33+1
Step 3: Equate the exponents and solve for x.
x+ 1 = 3 + 1
x+ 1 = 4
x= 4 −1
x= 3
Step 4: Check the solution by substituting x= 3 back into the original
equation. 1
2·33+1 = 27
1
2·34= 27
1
2·81 = 27
40.5 = 27
Step 5: Since the last step yielded a false statement, there is no solution to
the equation 1
2·3x+1 = 27.
Question 26
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
We know that 27 = 33. So, we can rewrite the equation as 32x−1= 33.
Step 2: Set the exponents equal to each other.
Since the bases are the same, we can set the exponents equal to each other:
2x−1 = 3.
Step 3: Solve for x.
Add 1to both sides of the equation to isolate 2x:2x= 4.
Divide by 2to solve for x:x= 2.
Step 4: Check the solution.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
Simplify both sides: 33= 33.
Since the left side is equal to the right side, the solution x= 2 is verified.
17
Question 27
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation 32x−1= 27:
32x−1= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
2x−1 = 3
Step 4: Solve for xby adding 1to both sides and then dividing by 2:
2x−1 = 3
2x= 4
x= 2
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 28
Question
Solve for x:2x+3 −4·2x+1 + 4 = 0.
Solution
Step 1: Let’s rewrite the equation in terms of 2xto make the manipulation
easier. Let y= 2x. Then the equation becomes:
y2−4y+ 4 = 0
Step 2: Now we need to solve the quadratic equation y2−4y+ 4 = 0.
Step 3: Factoring the quadratic, we get (y−2)2= 0.
Step 4: Taking the square root of both sides, we find y−2 = 0, so y= 2.
Step 5: Recall that y= 2x, so 2x= 2.
Step 6: Solving for x, we find x= 1.
Therefore, the solution to the equation 2x+3 −4·2x+1 + 4 = 0 is x= 1.
18
Question 29
Question
Solve the following exponential equation for x:3x−1−3x= 3.
Solution
Step 1: Let’s start by rewriting the equation using a common base. We know
that 3x−1=3x
3. Substituting this into the equation, we get 3x
3−3x= 3.
Step 2: Next, let’s combine the fractions on the left side of the equation. We
have 3x
−3·3x
3= 3.
Step 3: Simplifying further gives us −2·3x
3= 3.
Step 4: Multiplying both sides by 3, we get −2·3x= 9.
Step 5: Now, divide by -2 to solve for 3x. We have 3x=−9
2.
Step 6: Since 3xshould be a positive real number, there is no real solution
for xin this case. Therefore, the equation 3x−1−3x= 3 has no real solution.
Question 30
Question
Samantha invests $5000 in an account that earns 3.5% interest compounded
continuously. How long will it take for Samantha’s investment to double?
Solution
Step 1: Determine the formula to calculate the amount in the account after
time tyears with continuous compounding. The formula is given by:
A=P·ert
where: A= amount in the account after tyears, P= initial investment, r
= interest rate per year, t= time in years, and eis the base of the natural
logarithm, approximately equal to 2.71828.
Step 2: For this scenario, Samantha wants her investment to double, so the
amount in the account (A) should be equal to twice the initial investment (2P):
2P=P·e0.035t
Step 3: Divide both sides of the equation by Pto simplify:
2 = e0.035t
Step 4: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.035t)
19
ln(2) = 0.035tln(e)
ln(2) = 0.035t
Step 5: Divide by 0.035 to solve for t:
t=ln(2)
0.035 ≈19.81 years
Therefore, it will take approximately 19.81 years for Samantha’s investment
to double in the account with continuous compounding.
Question 31
Question
A bacteria culture initially contains 100 bacteria and doubles in size every 2
hours.
(a) Find a formula for the number of bacteria after t hours.
(b) How many bacteria will be present after 6 hours?
Solution
(a) Let N(t)represent the number of bacteria after thours. Since the bacteria
culture doubles in size every 2 hours, we have:
N(t) = 100 ·2t
2
(b) To find the number of bacteria after 6 hours, we substitute t= 6 into
the formula we found in part (a):
N(6) = 100 ·26
2
N(6) = 100 ·23
N(6) = 100 ·8
N(6) = 800
After 6 hours, there will be 800 bacteria present in the culture.
Question 32
Question
Samantha invests $5000 in a savings account that earns 4% interest compounded
continuously. How long will it take for her money to double?
20
Solution
Step 1: The formula for continuously compounded interest is given by A=P ert,
where: - Ais the amount of money accumulated after tyears, - Pis the principal
amount (initial investment), - ris the annual interest rate in decimal form, - t
is the time the money is invested for, - eis Euler’s number (≈2.71828).
Step 2: In this case, Samantha invested P= $5000, the interest rate is
r= 0.04, and she wants her money to double. Therefore, the amount she wants
is 2P= $10000.
Step 3: Plugging in the given values into the continuously compounded
interest formula, we have:
10000 = 5000 ·e0.04t
Step 4: Divide both sides by 5000:
2 = e0.04t
Step 5: To solve for t, we take the natural logarithm of both sides:
ln(2) = ln(e0.04t)
Step 6: Remember that ln(ex) = x, so we get:
ln(2) = 0.04t
Step 7: Now, isolate tby dividing by 0.04:
t=ln(2)
0.04
Step 8: Using a calculator, we find:
t≈0.693147
0.04 ≈17.33
Step 9: Therefore, it will take approximately 17.33 years for Samantha’s
investment to double.
Question 33
Question
Solve the exponential equation 32x+1 = 27.
21
Solution
Step 1: Rewrite 27 as a power of 3:27 = 33.
Step 2: Substitute 33into the original equation: 32x+1 = 33.
Step 3: Since the bases are the same, we can set the exponents equal to each
other: 2x+ 1 = 3.
Step 4: Solve for xby isolating the variable:
2x+ 1 = 3
2x= 3 −1
2x= 2
x=2
2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation:
32(1)+1 = 27
33= 27
27 = 27
Since the solution satisfies the original equation, x= 1 is the correct solution.
Question 34
Question
Solve the exponential equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: We can rewrite 27 as 33. Therefore, the equation becomes:
3x−1= 33
Step 2: Since the base is the same, we can set the exponents equal to each
other:
x−1 = 3
Now, solve for x:
x= 3 + 1
x= 4
Therefore, the solution to the equation 3x−1= 27 is x= 4.
22
Question 35
Question
A certain investment grows according to the formula A(t) = 5000 ·e0.04t, where
A(t)represents the amount of money in the account after tyears. Calculate the
time it will take for the investment to double.
Solution
Step 1: To find when the investment doubles, we need to set A(t)equal to twice
the initial amount:
2·5000 = 5000 ·e0.04t
Step 2: Simplify the equation:
10000 = 5000 ·e0.04t
Step 3: Divide both sides by 5000 to isolate the exponential term:
2 = e0.04t
Step 4: To solve for t, take the natural logarithm of both sides to undo the
exponential:
ln(2) = ln(e0.04t)
Step 5: By the property of logarithms, we can bring down the exponent:
ln(2) = 0.04tln(e)
Step 6: Since ln(e) = 1, the equation becomes:
ln(2) = 0.04t
Step 7: Finally, divide by 0.04 to solve for t:
t=ln(2)
0.04 ≈0.6931
0.04 ≈17.33 years
Therefore, it will take approximately 17.33 years for the investment to dou-
ble.
23
