MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 4
Liberty University
Question 1
Question
Samantha deposits $5000 into a savings account that offers a 5% annual interest
rate compounded continuously. How much will Samantha have in her account
after 10 years?
Solution
Step 1: The formula for compound interest compounded continuously is given
by
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (the initial amount of money). - 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: In this case, P= 5000,r= 0.05, and t= 10. Plugging these values
into the formula gives:
A= 5000 ·e0.05·10
Step 3: Calculating the exponential term:
e0.05·10 =e0.5
Step 4: Evaluating e0.5using a calculator gives approximately 1.64872.
Step 5: Substituting this back into the original formula:
A= 5000 ·1.64872
Step 6: Multiplying gives:
A= 8243.6
Therefore, after 10 years, Samantha will have approximately $8243.60 in her
savings account.
Question 2
Question
Sara has invested $5000 in a savings account that grows continuously at an
annual interest rate of 6%. Write an exponential model to represent the balance
Bafter tyears. How long will it take for her investment to double in value?
Solution
Step 1: The continuous growth formula is given by B=P·ert, where Pis the
principal amount, ris the interest rate per time unit, tis the time in years, and
Bis the balance after tyears. Here, P= 5000,r= 0.06, and we need to find
an expression for tthat represents the time taken for the balance to double.
Step 2: Substituting P= 5000 and r= 0.06 into the continuous growth
formula, we get B= 5000 ·e0.06t.
Step 3: We want to find the time it takes for the initial investment to
double. This means we are looking for the value of tthat satisfies the equation
2·5000 = 5000 ·e0.06t.
Step 4: Solving the equation 2·5000 = 5000 ·e0.06tfor t:
10000 = 5000 ·e0.06t
Step 5: Divide both sides by 5000:
2 = e0.06t
Step 6: Take the natural logarithm of both sides to solve for t:
ln(2) = lne0.06t
Step 7: Using the property of logarithms that ln(ex) = x, we have:
ln(2) = 0.06t
Step 8: Solve for tby dividing by 0.06:
t=ln(2)
0.06 ≈11.57 years
Therefore, it will take approximately 11.57 years for Sara’s investment to
double in value.
2
Question 3
Question
Solve the following exponential equation for x:3x+1 −10 ·3x+ 24 = 0.
Solution
Step 1: Let’s rewrite the equation in a way that will help us solve for x.
3x+1 −10 ·3x+ 24 = 0
3·3x−10 ·3x+ 24 = 0
3x(3 −10) + 24 = 0
3x(−7) + 24 = 0
3x(−7) = −24
Step 2: Now, we can rewrite the equation as an exponential with the same
base:
3x=−24
−7
Step 3: Simplify the right-hand side:
3x=24
7
Step 4: Taking the logarithm of both sides with base 3:
log3(3x) = log324
7
x·log3(3) = log324
7
x= log324
7
Therefore, the solution to the given exponential equation is x= log324
7.
Question 4
Question
Solve the exponential equation for x: 32x= 27.
3
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Set the equation as 32x= 33and equate the exponents:
2x= 3
Step 3: Solve for x by dividing both sides by 2:
x=3
2= 1.5
Therefore, the solution to the exponential equation 32x= 27 is x= 1.5.
Question 5
Question
Solve the exponential equation 23x+4 = 32 for x.
Solution
Step 1: Rewrite both sides of the equation with the same base.
23x+4 = 32
23x+4 = 25
Step 2: Set the exponents equal to each other.
3x+ 4 = 5
Step 3: Solve for x.
3x+ 4 = 5
3x= 1
x=1
3
Therefore, the solution to the equation 23x+4 = 32 is x=1
3.
Question 6
Question
Suppose a population of bacteria doubles every hour. If there are initially 100
bacteria, how many bacteria will there be after 6 hours?
4
Solution
Step 1: To model the population growth, we can use the exponential function
P(t) = P0·2t, where P(t)is the population at time t,P0is the initial population
(100 bacteria in this case), and tis the time in hours.
Step 2: Substituting P0= 100 and t= 6 into the exponential function, we
get:
P(6) = 100 ·26
Step 3: Calculating the expression gives:
P(6) = 100 ·64 = 6400
Step 4: Therefore, after 6 hours, there will be 6400 bacteria in the popula-
tion.
Question 7
Question
Solve the equation 4x−1= 8 for x.
Solution
Step 1: Rewrite the equation in terms of the base 4.
4x−1= 8
4x−1= 43
2
Step 2: Since the bases are the same, we can set the exponents equal to each
other.
x−1 = 3
2
Step 3: Solve for x.
x=3
2+ 1
Step 4: Simplify.
x=3
2+2
2=5
2
Step 5: Therefore, the solution to the equation 4x−1= 8 is x=5
2.
Question 8
Question
Samantha invested $10,000 in a savings account that pays 3% interest com-
pounded quarterly. How much money will Samantha have in her account after
5 years?
5
Solution
Step 1: First, we need to determine the variables involved in the compound
interest formula. The formula for compound interest is given by:
A=P1 + r
nnt
where: - Ais the amount of money accumulated after tyears, including in-
terest, - Pis the principal amount (the initial amount of money), - ris the
annual interest rate (in decimal form), - nis the number of times that interest
is compounded per year, and - tis the time the money is invested for in years.
In this case, we have: - P= $10,000 (the initial investment), - r= 0.03 (3%
interest rate in decimal form), - n= 4 (quarterly compounding), and - t= 5 (5
years).
Step 2: Substitute the given values into the compound interest formula and
solve for A.
A= 10000 1 + 0.03
44∗5
Step 3: Calculate the amount in Samantha’s account after 5 years.
A= 10000 1 + 0.03
420
A= 10000 (1 + 0.0075)20
A= 10000 ×1.007520
A= 10000 ×1.161067
Therefore, Samantha will have approximately $11,610.67 in her account after
5 years.
Question 9
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite the equation in terms of the same base:
32x−1= 33
Step 2: Set the exponents equal to each other:
2x−1 = 3
6
Step 3: Solve for x:
2x= 4
x= 2
Step 4: Check the solution in the original equation:
32(2)−1= 33
34−1= 33
33= 33
27 = 27
Therefore, the solution to the equation is x= 2.
Question 10
Question
Suppose a bacteria culture has an initial population of 1000 and the population
doubles every 6 hours. Write an exponential function to model the population
of the bacteria culture after thours.
Solution
Step 1: Let P(t)represent the population of the bacteria culture after thours.
Since the population doubles every 6 hours, we have the growth factor r= 2.
Step 2: To find the initial population P0, we use the given information that
the initial population is 1000. Therefore, P0= 1000.
Step 3: The exponential function that models the population growth can be
written as:
P(t) = P0×(r)t
time for doubling
Step 4: Substituting P0= 1000,r= 2, and time for doubling = 6 hours into
the equation, we have:
P(t) = 1000 ×2t
6
Therefore, the exponential function that models the population of the bac-
teria culture after thours is P(t) = 1000 ×2t
6.
Question 11
Question
The population of a city is initially 500,000 people, and it is projected to grow
exponentially at a rate of 3% per year. Find a function that models the popu-
lation P(t)after tyears.
7
Solution
Step 1: Recall that the formula for exponential growth is given by P(t) =
P0·(1 + r)t, where: - P(t)is the population after tyears, - P0is the initial
population, - ris the growth rate as a decimal, and - tis the number of years.
Step 2: Substitute the given values into the formula: P(t) = 500,000 ·(1 +
0.03)t
Step 3: Simplify the expression: P(t) = 500,000 ·(1.03)t
Therefore, the function that models the population P(t)after tyears is given
by P(t) = 500,000 ·(1.03)t.
Question 12
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 in terms of a base of 3.
27 = 33
Step 2: Substitute the new representation of 27 into the 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 x.
2x−1 = 3
2x= 4
x= 2
Step 5: Check the solution by substituting x= 2 back into the original
equation.
32(2)−1= 27
33= 27
27 = 27
Therefore, the solution to the exponential equation is x= 2.
8
Question 13
Question
Consider the function f(x) = 2x−3. Find the inverse function f−1(x)and
evaluate f−1(5).
Solution
Step 1: To find the inverse function f−1(x), we start by swapping xand yand
then solve for y.
x= 2y−3
Step 2: Add 3 to both sides to isolate 2y.
x+ 3 = 2y
Step 3: Take the logarithm of both sides to solve for y.
log(x+ 3) = log(2y)
log(x+ 3) = ylog(2)
y=log(x+ 3)
log(2)
Step 4: Therefore, the inverse function f−1(x)is:
f−1(x) = log(x+ 3)
log(2)
Step 5: To evaluate f−1(5), substitute x= 5 into the inverse function.
f−1(5) = log(5 + 3)
log(2)
f−1(5) = log(8)
log(2)
f−1(5) = 3 log(2)
log(2)
f−1(5) = 3
Therefore, f−1(5) = 3.
Question 14
Question
Samantha invested $10,000 in a savings account that earns 4% interest com-
pounded quarterly. How long will it take for her money to double?
9
Solution
Step 1: The formula for compound interest is given by:
A=P1 + r
nnt
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: In this case, Samantha’s initial investment is $10,000, the annual
interest rate is 4%, and interest is compounded quarterly (so n= 4). Therefore,
we have:
2·10,000 = 10,000 1 + 0.04
44t
Step 3: Simplify the equation:
20,000 = 10,000 (1 + 0.01)4t
2 = (1.01)4t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln (1.01)4t
ln(2) = 4tln(1.01)
Step 5: Solve for t:
t=ln(2)
4 ln(1.01)
Step 6: Calculate the value of tusing a calculator:
t≈0.6931
4·0.0099
t≈0.6931
0.0396
t≈17.5
Step 7: It will take approximately 17.5 years for Samantha’s money to double
in the savings account.
Question 15
Question
Consider the exponential function f(x) = 3 ·2x.
If the population of a town is modeled by this function, where xrepresents
the number of years since the year 2020, determine the population of the town
in the year 2050. Round your answer to the nearest whole number.
10
Solution
Step 1: To find the population of the town in the year 2050, we need to determine
the value of f(2050) by substituting x= 2050 −2020 = 30 into the function
f(x) = 3 ·2x.
Step 2: Substitute x= 30 into the function f(x):
f(30) = 3 ·230
Step 3: Calculate 230:
230 = 1073741824
Step 4: Substitute 230 = 1073741824 into the expression for f(30):
f(30) = 3 ·1073741824
Step 5: Calculate the population of the town in the year 2050 by multiplying
3 by 1073741824:
f(30) = 3221225472
Therefore, the population of the town in the year 2050 is approximately
3,221,225,472.
Question 16
Question
Solve the exponential equation 32x+ 3x−10 = 0 for x.
Solution
Step 1: Let u= 3x. Then our equation becomes u2+u−10 = 0.
Step 2: Factor the quadratic equation u2+u−10 = 0 to get (u+2)(u−5) = 0.
Step 3: Set each factor equal to zero and solve for u:
u+ 2 = 0 or u−5 = 0
Step 4: Solve the first equation u+ 2 = 0 to get u=−2.
Step 5: Solve the second equation u−5 = 0 to get u= 5.
Step 6: Recall that u= 3x. Substitute these values back to solve for x:
3x=−2or 3x= 5
Step 7: Neither of these equations have a real solution as an exponential
function is always positive. Therefore, the original equation 32x+ 3x−10 = 0
has no real solution.
11
Question 17
Question
Samantha invests $5000 in an account that earns 3.5% interest compounded
continuously. How long will it take for her money to double?
Solution
Step 1: We will use the continuous compound interest formula A=P·ert,
where: - Ais the amount of money after a certain time - Pis the principal
amount (initial investment) - ris the interest rate - tis the time in years
Step 2: Since Samantha wants her money to double, the amount of money
after doubling will be 2P. So, we can set up the equation as:
2P=P·e0.035t
Step 3: Now, we can solve for t.
e0.035t= 2
0.035t= ln(2)
t=ln(2)
0.035
Step 4: Calculating the value of tusing a calculator:
t≈0.6931
0.035 ≈19.802
Step 5: It will take approximately 19.802 years for Samantha’s money to
double in the account.
Question 18
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,we have 32x+1 = 33.
Step 2: Set the exponents equal to each other.
Setting the exponents equal to each other gives us 2x+ 1 = 3.
12
Step 3: Solve for x.
2x+ 1 = 3
2x= 2
x= 1.
Step 4: Check the solution.
We substitute x= 1 back into the original equation to check for extraneous solutions.
32(1)+1 = 27
33= 27
27 = 27 ✓
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 19
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4%, compounded continuously. How long will it take for her investment to
double?
Solution
Step 1: We can model the amount of money in Samantha’s account after tyears
using the formula for continuous compounding:
A(t) = P·ert
where Pis the principal amount (initial investment), ris the annual interest
rate (in decimal form), and tis time in years.
Step 2: The amount we want to calculate is when her investment doubles,
so we want to find tsuch that A(t)=2P. Substituting in the given values, we
have:
2P=P·e0.04t
2 = e0.04t
Step 3: To solve for t, we need to isolate the variable in the exponential
equation. Taking the natural logarithm of both sides will help us do this:
ln 2 = ln e0.04t
ln 2 = 0.04tln e
13
ln 2 = 0.04t
Step 4: Now, we can solve for tby dividing both sides by 0.04:
t=ln 2
0.04 ≈0.6931
0.04 ≈17.33 years
Therefore, it will take approximately 17.33 years for Samantha’s investment
to double in value.
Question 20
Question
Solve for xin the equation 23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Since 8 = 23, we have:
23x−1= 23
Step 2: Set the exponents equal to each other. Since the bases are the same,
we can set the exponents equal to each other:
3x−1 = 3
Step 3: Solve for x. Add 1 to both sides of the equation:
3x= 4
Step 4: Divide by 3 to solve for x.
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
Question 21
Question
A certain radioactive substance has a half-life of 10 days. If there are initially
200 grams of the substance, find an exponential function that models the amount
of the substance remaining after tdays.
14
Solution
Step 1: The general formula for exponential decay is given by A(t) = A0·(0.5) t
h,
where: A(t)is the amount of substance remaining after tdays, A0is the initial
amount of substance, and his the half-life of the substance.
Step 2: Substituting A0= 200 grams and h= 10 days into the formula, we
get A(t) = 200 ·(0.5) t
10 .
Therefore, the exponential function that models the amount of the substance
remaining after tdays is A(t) = 200 ·(0.5) t
10 .
Question 22
Question
Solve the exponential equation: 3x−5·3x−1+ 6 = 0.
Solution
Step 1: Notice that the given equation can be written as a quadratic equation
in terms of 3x. Let y= 3x, then the equation becomes y2−5y+ 6 = 0.
Step 2: Now, we need to solve the quadratic equation y2−5y+ 6 = 0. The
factored form is (y−2)(y−3) = 0.
Step 3: Setting each factor to zero gives y−2 = 0 or y−3 = 0.
Step 4: Solve for yin each case. y−2 = 0 ⇒y= 2 and y−3 = 0 ⇒y= 3.
Step 5: Recall that y= 3x. So, 3x= 2 or 3x= 3.
Step 6: Solve for xin each case. For 3x= 2, we have x= log3(2). For
3x= 3, we have x= 1.
Therefore, the solutions to the original equation 3x−5·3x−1+ 6 = 0 are
x= log3(2) and x= 1.
Question 23
Question
Solve the exponential equation 32x+1 = 81.
Solution
Step 1: Rewrite 81 as a power of 3.
81 = 34
Step 2: Set the equation equal to the rewritten form of 81.
32x+1 = 34
15
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 4
Step 4: Solve for x.
2x= 4 −1
2x= 3
x=3
2
x= 1.5
Step 5: Check the solution.
32(1.5)+1 = 34
33+1 = 34
34= 34
Since both sides are equal, the solution x= 1.5is correct.
Question 24
Question
Samantha invested $10,000 in a savings account that earns 4% annual interest
compounded continuously. How much will the investment be worth after 10
years?
Solution
Step 1: We can use the formula for continuous compounding:
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: In this case, P= $10,000,r= 0.04, and t= 10 years. Substituting
these values into the formula, we get:
A= 10000 ·e0.04·10.
Step 3: Simplifying further, we get:
A= 10000 ·e0.4.
16
Step 4: We can use the approximate value of e≈2.718 to approximate the
final amount:
A≈10000 ·2.7180.4.
Step 5: Calculating the exponential term, we get:
A≈10000 ·1.4918.
Step 6: Finally, we find:
A≈$14,918.
Therefore, after 10 years, Samantha’s investment will be worth approxi-
mately $14,918.
Question 25
Question
Let f(x) = 4(3)x. Find the inverse function f−1(x).
Solution
Step 1: Replace f(x)with yto get y= 4(3)x.
Step 2: Switch xand yto get x= 4(3)y.
Step 3: Solve for yby isolating (3)y:
x= 4(3)y
x
4= (3)y
Step 4: Take the natural logarithm of both sides to bring down the exponent:
ln x
4= ln ((3)y)
ln x
4=yln(3)
Step 5: Divide by ln(3) to solve for y:
y=ln x
4
ln(3)
Step 6: Replace ywith f−1(x)to get the inverse function:
f−1(x) = ln x
4
ln(3)
17
Question 26
Question
Solve the exponential equation 2x+1 −4·2x+ 4 = 0.
Solution
Step 1: Let’s rewrite the equation in terms of a single base (2x) to simplify the
calculations.
2·2x−4·2x+ 4 = 0
Step 2: Combining like terms, we have
2x−4·2x+ 4 = 0
Step 3: Factor out 2xfrom the equation to get
2x(1 −4) + 4 = 0
Step 4: Simplify further to obtain
−2x+ 4 = 0
Step 5: Move 4 to the other side to isolate 2xto get
2x= 4
Step 6: Rewrite the equation using the same base to obtain
22= 2x
Step 7: Since the bases are the same, we can equate the exponents to find x
2 = x
Step 8: Therefore, the solution to the exponential equation 2x+1−4·2x+4 = 0
is x= 2.
Question 27
Question
Samantha invested $10,000 in a savings account that pays 3
18
Solution
Step 1: First, we need to determine the interest rate per compounding period.
Since the interest is compounded monthly at a rate of 3
Step 2: Next, we use the formula for compound interest:
A=P1 + r
nnt
where: - Ais the amount of money accumulated after tyears, - Pis the
principal amount (initial investment), - ris the annual interest rate (as a deci-
mal), - nis the number of times that interest is compounded per year, and - t
is the time the money is invested for in years.
Step 3: Substitute the given values into the formula:
A= 10000 1 + 0.03
12 12·5
Step 4: Simplify the expression inside the parentheses:
A= 10000 (1 + 0.0025)60
Step 5: Calculate the value inside the parentheses:
A= 10000(1.0025)60
Step 6: Finally, compute the final amount:
A≈10000 ×1.1611778 ≈$11,611.78
Therefore, after 5 years, Samantha will have approximately $11,611.78 in
the savings account.
Question 28
Question
Solve the following exponential equation for x:52x−1= 25.
Solution
Step 1: Rewrite 25 as a power of 5.
25 = 52
Step 2: Substitute 52into the equation and solve for x.
52x−1= 52
19
Step 3: Since the bases are the same, set the exponents equal to each other.
2x−1 = 2
Step 4: Solve the equation for x.
2x−1 = 2
2x= 3
x=3
2
Step 5: Check the solution by substituting x=3
2back into the original
equation.
52( 3
2)−1= 52
53−1= 52
52= 52
Step 6: Since the left side equals the right side, the solution is x=3
2.
Question 29
Question
Solve the exponential equation 3x+2 = 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33, we can rewrite the exponential equation as
3x+2 = 33
.
Step 2: Solve for x.
Since both sides have the same base (3), we can equate the exponents:
x+ 2 = 3.
Now, solve for x:
x= 3 −2
x= 1.
Therefore, the solution to the exponential equation 3x+2 = 27 is x= 1.
20
Question 30
Question
Suppose the population of a small town is modeled by the function P(t) = 5000·
1.02t, where trepresents the number of years since the start of the observation.
Find the rate at which the population is growing after 5 years.
Solution
Step 1: Calculate the derivative of the population function P(t)with respect to
time t.d
dt P(t) = d
dt (5000 ·1.02t)
Step 2: Use the exponential function rule d
dt (at) = at·ln(a)and apply it to
the function 1.02t.d
dt P(t) = 5000 ·1.02t·ln(1.02)
Step 3: Evaluate the rate at which the population is growing after 5 years
by substituting t= 5 into the expression we found in Step 2.
d
dt P(t)
t=5
= 5000 ·1.025·ln(1.02)
Step 4: Perform the calculations to find the rate of population growth after
5 years.
d
dt P(t)
t=5
= 5000 ·1.10408 ≈5520.4
Therefore, the rate at which the population is growing after 5 years is ap-
proximately 5520.4 individuals per year.
Question 31
Question
An investment of 5000ismadeinanaccountthatpays4
Solution
Let A(t)be the amount of money in the account after tyears. The formula for
exponential growth with continuous compounding is given by:
A(t) = P ert,
where: - Pis the initial investment ($5000), - ris the interest rate (4- tis the
time in years, and - A(t)is the amount after tyears.
21
We are looking for the time tit takes for the investment to double, so we
want A(t)to be twice the initial investment amount. Thus, we want to solve
the equation 2P=P ert for t.
Step 1: Plug in the given values and set up the equation.
2(5000) = 5000e0.04t
Step 2: Simplify the equation.
10000 = 5000e0.04t
Step 3: Divide by 5000 to isolate the exponential term.
2 = e0.04t
Step 4: Take the natural logarithm of both sides to solve for t.
ln(2) = lne0.04t
Step 5: Use the property ln(ex) = x.
ln(2) = 0.04t
Step 6: Solve for t.
t=ln(2)
0.04
Step 7: Calculate the approximate value of t.
t≈0.6931
0.04 ≈17.33 years
So, it will take approximately 17.33 years for the investment to double in
value.
Question 32
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the equation.
32x−1= 33
22
Step 3: Since the bases are equal, we can equate the exponents.
2x−1 = 3
Step 4: Solve for x.
2x= 4
x= 2
Step 5: Check the solution.
32(2)−1= 33
33= 27
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 33
Question
Solve the exponential equation 32x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,we have 32x= 33.
Step 2: Equate the exponents.
Setting the exponents equal to each other, we get 2x= 3.
Step 3: Solve for x.
Dividing both sides by 2, we find x=3
2.
Therefore, the solution to the exponential equation 32x= 27 is x=3
2.
Question 34
Question
Suppose a population of bacteria doubles every 6 hours. If the initial population
is 1000 bacteria, find an exponential model for the population P(t)in terms of
time tin hours.
23
Solution
Step 1: Determine the growth factor
Since the population doubles every 6 hours, the growth factor ris 2.
Step 2: Write the exponential model
The exponential model for population growth is given by the formula:
P(t) = P0·(1 + r)t
k
where - P(t)is the population at time t, - P0is the initial population, - ris the
growth factor, and - kis the time it takes for the population to double.
Plugging in the values we know: - P0= 1000 (initial population) - r= 2
(growth factor) - k= 6 (time for population to double)
We have:
P(t) = 1000 ·(1 + 2) t
6
Step 3: Simplify the model
Simplify the expression to get the exponential model:
P(t) = 1000 ·2t
6
Therefore, the exponential model for the population P(t)in terms of time t
is P(t) = 1000 ·2t
6.
Question 35
Question
Let y= 23x−2x. Find the x-coordinate of the vertex for the graph of the
function y.
Solution
Step 1: Rewrite the function yin terms of a single exponential term. Step 2:
Identify the general form of an exponential function y=ax. Step 3: Find the
x-coordinate of the vertex using the formula x=−b
2a.
Step 1: Rewrite the function yin terms of a single exponential term.
Given y= 23x−2x, we can rewrite it as y= (2x)3−2x.
Step 2: Identify the general form of an exponential function y=ax.
We can rewrite the function in the form y=axby letting a= 2x. Thus,
y=a3−a.
Step 3: Find the x-coordinate of the vertex using the formula x=−b
2a.
The x-coordinate of the vertex of the function y=a3−ais given by x=
−−1
2(1) =1
2.
Therefore, the x-coordinate of the vertex for the graph of the function yis
1
2.
24
Question 3
Question
Solve the following exponential equation for x:3x+1 −10 ·3x+ 24 = 0.
Solution
Step 1: Let’s rewrite the equation in a way that will help us solve for x.
3x+1 −10 ·3x+ 24 = 0
3·3x−10 ·3x+ 24 = 0
3x(3 −10) + 24 = 0
3x(−7) + 24 = 0
3x(−7) = −24
Step 2: Now, we can rewrite the equation as an exponential with the same
base:
3x=−24
−7
Step 3: Simplify the right-hand side:
3x=24
7
Step 4: Taking the logarithm of both sides with base 3:
log3(3x) = log324
7
x·log3(3) = log324
7
x= log324
7
Therefore, the solution to the given exponential equation is x= log324
7.
Question 4
Question
Solve the exponential equation for x: 32x= 27.
3
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Set the equation as 32x= 33and equate the exponents:
2x= 3
Step 3: Solve for x by dividing both sides by 2:
x=3
2= 1.5
Therefore, the solution to the exponential equation 32x= 27 is x= 1.5.
Question 5
Question
Solve the exponential equation 23x+4 = 32 for x.
Solution
Step 1: Rewrite both sides of the equation with the same base.
23x+4 = 32
23x+4 = 25
Step 2: Set the exponents equal to each other.
3x+ 4 = 5
Step 3: Solve for x.
3x+ 4 = 5
3x= 1
x=1
3
Therefore, the solution to the equation 23x+4 = 32 is x=1
3.
Question 6
Question
Suppose a population of bacteria doubles every hour. If there are initially 100
bacteria, how many bacteria will there be after 6 hours?
4
Solution
Step 1: To model the population growth, we can use the exponential function
P(t) = P0·2t, where P(t)is the population at time t,P0is the initial population
(100 bacteria in this case), and tis the time in hours.
Step 2: Substituting P0= 100 and t= 6 into the exponential function, we
get:
P(6) = 100 ·26
Step 3: Calculating the expression gives:
P(6) = 100 ·64 = 6400
Step 4: Therefore, after 6 hours, there will be 6400 bacteria in the popula-
tion.
Question 7
Question
Solve the equation 4x−1= 8 for x.
Solution
Step 1: Rewrite the equation in terms of the base 4.
4x−1= 8
4x−1= 43
2
Step 2: Since the bases are the same, we can set the exponents equal to each
other.
x−1 = 3
2
Step 3: Solve for x.
x=3
2+ 1
Step 4: Simplify.
x=3
2+2
2=5
2
Step 5: Therefore, the solution to the equation 4x−1= 8 is x=5
2.
Question 8
Question
Samantha invested $10,000 in a savings account that pays 3% interest com-
pounded quarterly. How much money will Samantha have in her account after
5 years?
5
Solution
Step 1: First, we need to determine the variables involved in the compound
interest formula. The formula for compound interest is given by:
A=P1 + r
nnt
where: - Ais the amount of money accumulated after tyears, including in-
terest, - Pis the principal amount (the initial amount of money), - ris the
annual interest rate (in decimal form), - nis the number of times that interest
is compounded per year, and - tis the time the money is invested for in years.
In this case, we have: - P= $10,000 (the initial investment), - r= 0.03 (3%
interest rate in decimal form), - n= 4 (quarterly compounding), and - t= 5 (5
years).
Step 2: Substitute the given values into the compound interest formula and
solve for A.
A= 10000 1 + 0.03
44∗5
Step 3: Calculate the amount in Samantha’s account after 5 years.
A= 10000 1 + 0.03
420
A= 10000 (1 + 0.0075)20
A= 10000 ×1.007520
A= 10000 ×1.161067
Therefore, Samantha will have approximately $11,610.67 in her account after
5 years.
Question 9
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite the equation in terms of the same base:
32x−1= 33
Step 2: Set the exponents equal to each other:
2x−1 = 3
6
Step 3: Solve for x:
2x= 4
x= 2
Step 4: Check the solution in the original equation:
32(2)−1= 33
34−1= 33
33= 33
27 = 27
Therefore, the solution to the equation is x= 2.
Question 10
Question
Suppose a bacteria culture has an initial population of 1000 and the population
doubles every 6 hours. Write an exponential function to model the population
of the bacteria culture after thours.
Solution
Step 1: Let P(t)represent the population of the bacteria culture after thours.
Since the population doubles every 6 hours, we have the growth factor r= 2.
Step 2: To find the initial population P0, we use the given information that
the initial population is 1000. Therefore, P0= 1000.
Step 3: The exponential function that models the population growth can be
written as:
P(t) = P0×(r)t
time for doubling
Step 4: Substituting P0= 1000,r= 2, and time for doubling = 6 hours into
the equation, we have:
P(t) = 1000 ×2t
6
Therefore, the exponential function that models the population of the bac-
teria culture after thours is P(t) = 1000 ×2t
6.
Question 11
Question
The population of a city is initially 500,000 people, and it is projected to grow
exponentially at a rate of 3% per year. Find a function that models the popu-
lation P(t)after tyears.
7
Solution
Step 1: Recall that the formula for exponential growth is given by P(t) =
P0·(1 + r)t, where: - P(t)is the population after tyears, - P0is the initial
population, - ris the growth rate as a decimal, and - tis the number of years.
Step 2: Substitute the given values into the formula: P(t) = 500,000 ·(1 +
0.03)t
Step 3: Simplify the expression: P(t) = 500,000 ·(1.03)t
Therefore, the function that models the population P(t)after tyears is given
by P(t) = 500,000 ·(1.03)t.
Question 12
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 in terms of a base of 3.
27 = 33
Step 2: Substitute the new representation of 27 into the 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 x.
2x−1 = 3
2x= 4
x= 2
Step 5: Check the solution by substituting x= 2 back into the original
equation.
32(2)−1= 27
33= 27
27 = 27
Therefore, the solution to the exponential equation is x= 2.
8
Question 13
Question
Consider the function f(x) = 2x−3. Find the inverse function f−1(x)and
evaluate f−1(5).
Solution
Step 1: To find the inverse function f−1(x), we start by swapping xand yand
then solve for y.
x= 2y−3
Step 2: Add 3 to both sides to isolate 2y.
x+ 3 = 2y
Step 3: Take the logarithm of both sides to solve for y.
log(x+ 3) = log(2y)
log(x+ 3) = ylog(2)
y=log(x+ 3)
log(2)
Step 4: Therefore, the inverse function f−1(x)is:
f−1(x) = log(x+ 3)
log(2)
Step 5: To evaluate f−1(5), substitute x= 5 into the inverse function.
f−1(5) = log(5 + 3)
log(2)
f−1(5) = log(8)
log(2)
f−1(5) = 3 log(2)
log(2)
f−1(5) = 3
Therefore, f−1(5) = 3.
Question 14
Question
Samantha invested $10,000 in a savings account that earns 4% interest com-
pounded quarterly. How long will it take for her money to double?
9
Solution
Step 1: The formula for compound interest is given by:
A=P1 + r
nnt
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: In this case, Samantha’s initial investment is $10,000, the annual
interest rate is 4%, and interest is compounded quarterly (so n= 4). Therefore,
we have:
2·10,000 = 10,000 1 + 0.04
44t
Step 3: Simplify the equation:
20,000 = 10,000 (1 + 0.01)4t
2 = (1.01)4t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln (1.01)4t
ln(2) = 4tln(1.01)
Step 5: Solve for t:
t=ln(2)
4 ln(1.01)
Step 6: Calculate the value of tusing a calculator:
t≈0.6931
4·0.0099
t≈0.6931
0.0396
t≈17.5
Step 7: It will take approximately 17.5 years for Samantha’s money to double
in the savings account.
Question 15
Question
Consider the exponential function f(x) = 3 ·2x.
If the population of a town is modeled by this function, where xrepresents
the number of years since the year 2020, determine the population of the town
in the year 2050. Round your answer to the nearest whole number.
10
Solution
Step 1: To find the population of the town in the year 2050, we need to determine
the value of f(2050) by substituting x= 2050 −2020 = 30 into the function
f(x) = 3 ·2x.
Step 2: Substitute x= 30 into the function f(x):
f(30) = 3 ·230
Step 3: Calculate 230:
230 = 1073741824
Step 4: Substitute 230 = 1073741824 into the expression for f(30):
f(30) = 3 ·1073741824
Step 5: Calculate the population of the town in the year 2050 by multiplying
3 by 1073741824:
f(30) = 3221225472
Therefore, the population of the town in the year 2050 is approximately
3,221,225,472.
Question 16
Question
Solve the exponential equation 32x+ 3x−10 = 0 for x.
Solution
Step 1: Let u= 3x. Then our equation becomes u2+u−10 = 0.
Step 2: Factor the quadratic equation u2+u−10 = 0 to get (u+2)(u−5) = 0.
Step 3: Set each factor equal to zero and solve for u:
u+ 2 = 0 or u−5 = 0
Step 4: Solve the first equation u+ 2 = 0 to get u=−2.
Step 5: Solve the second equation u−5 = 0 to get u= 5.
Step 6: Recall that u= 3x. Substitute these values back to solve for x:
3x=−2or 3x= 5
Step 7: Neither of these equations have a real solution as an exponential
function is always positive. Therefore, the original equation 32x+ 3x−10 = 0
has no real solution.
11
Question 17
Question
Samantha invests $5000 in an account that earns 3.5% interest compounded
continuously. How long will it take for her money to double?
Solution
Step 1: We will use the continuous compound interest formula A=P·ert,
where: - Ais the amount of money after a certain time - Pis the principal
amount (initial investment) - ris the interest rate - tis the time in years
Step 2: Since Samantha wants her money to double, the amount of money
after doubling will be 2P. So, we can set up the equation as:
2P=P·e0.035t
Step 3: Now, we can solve for t.
e0.035t= 2
0.035t= ln(2)
t=ln(2)
0.035
Step 4: Calculating the value of tusing a calculator:
t≈0.6931
0.035 ≈19.802
Step 5: It will take approximately 19.802 years for Samantha’s money to
double in the account.
Question 18
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,we have 32x+1 = 33.
Step 2: Set the exponents equal to each other.
Setting the exponents equal to each other gives us 2x+ 1 = 3.
12
Step 3: Solve for x.
2x+ 1 = 3
2x= 2
x= 1.
Step 4: Check the solution.
We substitute x= 1 back into the original equation to check for extraneous solutions.
32(1)+1 = 27
33= 27
27 = 27 ✓
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 19
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4%, compounded continuously. How long will it take for her investment to
double?
Solution
Step 1: We can model the amount of money in Samantha’s account after tyears
using the formula for continuous compounding:
A(t) = P·ert
where Pis the principal amount (initial investment), ris the annual interest
rate (in decimal form), and tis time in years.
Step 2: The amount we want to calculate is when her investment doubles,
so we want to find tsuch that A(t)=2P. Substituting in the given values, we
have:
2P=P·e0.04t
2 = e0.04t
Step 3: To solve for t, we need to isolate the variable in the exponential
equation. Taking the natural logarithm of both sides will help us do this:
ln 2 = ln e0.04t
ln 2 = 0.04tln e
13
ln 2 = 0.04t
Step 4: Now, we can solve for tby dividing both sides by 0.04:
t=ln 2
0.04 ≈0.6931
0.04 ≈17.33 years
Therefore, it will take approximately 17.33 years for Samantha’s investment
to double in value.
Question 20
Question
Solve for xin the equation 23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Since 8 = 23, we have:
23x−1= 23
Step 2: Set the exponents equal to each other. Since the bases are the same,
we can set the exponents equal to each other:
3x−1 = 3
Step 3: Solve for x. Add 1 to both sides of the equation:
3x= 4
Step 4: Divide by 3 to solve for x.
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
Question 21
Question
A certain radioactive substance has a half-life of 10 days. If there are initially
200 grams of the substance, find an exponential function that models the amount
of the substance remaining after tdays.
14
Solution
Step 1: The general formula for exponential decay is given by A(t) = A0·(0.5) t
h,
where: A(t)is the amount of substance remaining after tdays, A0is the initial
amount of substance, and his the half-life of the substance.
Step 2: Substituting A0= 200 grams and h= 10 days into the formula, we
get A(t) = 200 ·(0.5) t
10 .
Therefore, the exponential function that models the amount of the substance
remaining after tdays is A(t) = 200 ·(0.5) t
10 .
Question 22
Question
Solve the exponential equation: 3x−5·3x−1+ 6 = 0.
Solution
Step 1: Notice that the given equation can be written as a quadratic equation
in terms of 3x. Let y= 3x, then the equation becomes y2−5y+ 6 = 0.
Step 2: Now, we need to solve the quadratic equation y2−5y+ 6 = 0. The
factored form is (y−2)(y−3) = 0.
Step 3: Setting each factor to zero gives y−2 = 0 or y−3 = 0.
Step 4: Solve for yin each case. y−2 = 0 ⇒y= 2 and y−3 = 0 ⇒y= 3.
Step 5: Recall that y= 3x. So, 3x= 2 or 3x= 3.
Step 6: Solve for xin each case. For 3x= 2, we have x= log3(2). For
3x= 3, we have x= 1.
Therefore, the solutions to the original equation 3x−5·3x−1+ 6 = 0 are
x= log3(2) and x= 1.
Question 23
Question
Solve the exponential equation 32x+1 = 81.
Solution
Step 1: Rewrite 81 as a power of 3.
81 = 34
Step 2: Set the equation equal to the rewritten form of 81.
32x+1 = 34
15
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 4
Step 4: Solve for x.
2x= 4 −1
2x= 3
x=3
2
x= 1.5
Step 5: Check the solution.
32(1.5)+1 = 34
33+1 = 34
34= 34
Since both sides are equal, the solution x= 1.5is correct.
Question 24
Question
Samantha invested $10,000 in a savings account that earns 4% annual interest
compounded continuously. How much will the investment be worth after 10
years?
Solution
Step 1: We can use the formula for continuous compounding:
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: In this case, P= $10,000,r= 0.04, and t= 10 years. Substituting
these values into the formula, we get:
A= 10000 ·e0.04·10.
Step 3: Simplifying further, we get:
A= 10000 ·e0.4.
16
Step 4: We can use the approximate value of e≈2.718 to approximate the
final amount:
A≈10000 ·2.7180.4.
Step 5: Calculating the exponential term, we get:
A≈10000 ·1.4918.
Step 6: Finally, we find:
A≈$14,918.
Therefore, after 10 years, Samantha’s investment will be worth approxi-
mately $14,918.
Question 25
Question
Let f(x) = 4(3)x. Find the inverse function f−1(x).
Solution
Step 1: Replace f(x)with yto get y= 4(3)x.
Step 2: Switch xand yto get x= 4(3)y.
Step 3: Solve for yby isolating (3)y:
x= 4(3)y
x
4= (3)y
Step 4: Take the natural logarithm of both sides to bring down the exponent:
ln x
4= ln ((3)y)
ln x
4=yln(3)
Step 5: Divide by ln(3) to solve for y:
y=ln x
4
ln(3)
Step 6: Replace ywith f−1(x)to get the inverse function:
f−1(x) = ln x
4
ln(3)
17
Question 26
Question
Solve the exponential equation 2x+1 −4·2x+ 4 = 0.
Solution
Step 1: Let’s rewrite the equation in terms of a single base (2x) to simplify the
calculations.
2·2x−4·2x+ 4 = 0
Step 2: Combining like terms, we have
2x−4·2x+ 4 = 0
Step 3: Factor out 2xfrom the equation to get
2x(1 −4) + 4 = 0
Step 4: Simplify further to obtain
−2x+ 4 = 0
Step 5: Move 4 to the other side to isolate 2xto get
2x= 4
Step 6: Rewrite the equation using the same base to obtain
22= 2x
Step 7: Since the bases are the same, we can equate the exponents to find x
2 = x
Step 8: Therefore, the solution to the exponential equation 2x+1−4·2x+4 = 0
is x= 2.
Question 27
Question
Samantha invested $10,000 in a savings account that pays 3
18
Solution
Step 1: First, we need to determine the interest rate per compounding period.
Since the interest is compounded monthly at a rate of 3
Step 2: Next, we use the formula for compound interest:
A=P1 + r
nnt
where: - Ais the amount of money accumulated after tyears, - Pis the
principal amount (initial investment), - ris the annual interest rate (as a deci-
mal), - nis the number of times that interest is compounded per year, and - t
is the time the money is invested for in years.
Step 3: Substitute the given values into the formula:
A= 10000 1 + 0.03
12 12·5
Step 4: Simplify the expression inside the parentheses:
A= 10000 (1 + 0.0025)60
Step 5: Calculate the value inside the parentheses:
A= 10000(1.0025)60
Step 6: Finally, compute the final amount:
A≈10000 ×1.1611778 ≈$11,611.78
Therefore, after 5 years, Samantha will have approximately $11,611.78 in
the savings account.
Question 28
Question
Solve the following exponential equation for x:52x−1= 25.
Solution
Step 1: Rewrite 25 as a power of 5.
25 = 52
Step 2: Substitute 52into the equation and solve for x.
52x−1= 52
19
Step 3: Since the bases are the same, set the exponents equal to each other.
2x−1 = 2
Step 4: Solve the equation for x.
2x−1 = 2
2x= 3
x=3
2
Step 5: Check the solution by substituting x=3
2back into the original
equation.
52( 3
2)−1= 52
53−1= 52
52= 52
Step 6: Since the left side equals the right side, the solution is x=3
2.
Question 29
Question
Solve the exponential equation 3x+2 = 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33, we can rewrite the exponential equation as
3x+2 = 33
.
Step 2: Solve for x.
Since both sides have the same base (3), we can equate the exponents:
x+ 2 = 3.
Now, solve for x:
x= 3 −2
x= 1.
Therefore, the solution to the exponential equation 3x+2 = 27 is x= 1.
20
Question 30
Question
Suppose the population of a small town is modeled by the function P(t) = 5000·
1.02t, where trepresents the number of years since the start of the observation.
Find the rate at which the population is growing after 5 years.
Solution
Step 1: Calculate the derivative of the population function P(t)with respect to
time t.d
dt P(t) = d
dt (5000 ·1.02t)
Step 2: Use the exponential function rule d
dt (at) = at·ln(a)and apply it to
the function 1.02t.d
dt P(t) = 5000 ·1.02t·ln(1.02)
Step 3: Evaluate the rate at which the population is growing after 5 years
by substituting t= 5 into the expression we found in Step 2.
d
dt P(t)
t=5
= 5000 ·1.025·ln(1.02)
Step 4: Perform the calculations to find the rate of population growth after
5 years.
d
dt P(t)
t=5
= 5000 ·1.10408 ≈5520.4
Therefore, the rate at which the population is growing after 5 years is ap-
proximately 5520.4 individuals per year.
Question 31
Question
An investment of 5000ismadeinanaccountthatpays4
Solution
Let A(t)be the amount of money in the account after tyears. The formula for
exponential growth with continuous compounding is given by:
A(t) = P ert,
where: - Pis the initial investment ($5000), - ris the interest rate (4- tis the
time in years, and - A(t)is the amount after tyears.
21
We are looking for the time tit takes for the investment to double, so we
want A(t)to be twice the initial investment amount. Thus, we want to solve
the equation 2P=P ert for t.
Step 1: Plug in the given values and set up the equation.
2(5000) = 5000e0.04t
Step 2: Simplify the equation.
10000 = 5000e0.04t
Step 3: Divide by 5000 to isolate the exponential term.
2 = e0.04t
Step 4: Take the natural logarithm of both sides to solve for t.
ln(2) = lne0.04t
Step 5: Use the property ln(ex) = x.
ln(2) = 0.04t
Step 6: Solve for t.
t=ln(2)
0.04
Step 7: Calculate the approximate value of t.
t≈0.6931
0.04 ≈17.33 years
So, it will take approximately 17.33 years for the investment to double in
value.
Question 32
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the equation.
32x−1= 33
22
Step 3: Since the bases are equal, we can equate the exponents.
2x−1 = 3
Step 4: Solve for x.
2x= 4
x= 2
Step 5: Check the solution.
32(2)−1= 33
33= 27
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 33
Question
Solve the exponential equation 32x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,we have 32x= 33.
Step 2: Equate the exponents.
Setting the exponents equal to each other, we get 2x= 3.
Step 3: Solve for x.
Dividing both sides by 2, we find x=3
2.
Therefore, the solution to the exponential equation 32x= 27 is x=3
2.
Question 34
Question
Suppose a population of bacteria doubles every 6 hours. If the initial population
is 1000 bacteria, find an exponential model for the population P(t)in terms of
time tin hours.
23
Solution
Step 1: Determine the growth factor
Since the population doubles every 6 hours, the growth factor ris 2.
Step 2: Write the exponential model
The exponential model for population growth is given by the formula:
P(t) = P0·(1 + r)t
k
where - P(t)is the population at time t, - P0is the initial population, - ris the
growth factor, and - kis the time it takes for the population to double.
Plugging in the values we know: - P0= 1000 (initial population) - r= 2
(growth factor) - k= 6 (time for population to double)
We have:
P(t) = 1000 ·(1 + 2) t
6
Step 3: Simplify the model
Simplify the expression to get the exponential model:
P(t) = 1000 ·2t
6
Therefore, the exponential model for the population P(t)in terms of time t
is P(t) = 1000 ·2t
6.
Question 35
Question
Let y= 23x−2x. Find the x-coordinate of the vertex for the graph of the
function y.
Solution
Step 1: Rewrite the function yin terms of a single exponential term. Step 2:
Identify the general form of an exponential function y=ax. Step 3: Find the
x-coordinate of the vertex using the formula x=−b
2a.
Step 1: Rewrite the function yin terms of a single exponential term.
Given y= 23x−2x, we can rewrite it as y= (2x)3−2x.
Step 2: Identify the general form of an exponential function y=ax.
We can rewrite the function in the form y=axby letting a= 2x. Thus,
y=a3−a.
Step 3: Find the x-coordinate of the vertex using the formula x=−b
2a.
The x-coordinate of the vertex of the function y=a3−ais given by x=
−−1
2(1) =1
2.
Therefore, the x-coordinate of the vertex for the graph of the function yis
1
2.
24
Question 3
Question
Solve the following exponential equation for x:3x+1 −10 ·3x+ 24 = 0.
Solution
Step 1: Let’s rewrite the equation in a way that will help us solve for x.
3x+1 −10 ·3x+ 24 = 0
3·3x−10 ·3x+ 24 = 0
3x(3 −10) + 24 = 0
3x(−7) + 24 = 0
3x(−7) = −24
Step 2: Now, we can rewrite the equation as an exponential with the same
base:
3x=−24
−7
Step 3: Simplify the right-hand side:
3x=24
7
Step 4: Taking the logarithm of both sides with base 3:
log3(3x) = log324
7
x·log3(3) = log324
7
x= log324
7
Therefore, the solution to the given exponential equation is x= log324
7.
Question 4
Question
Solve the exponential equation for x: 32x= 27.
3
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Set the equation as 32x= 33and equate the exponents:
2x= 3
Step 3: Solve for x by dividing both sides by 2:
x=3
2= 1.5
Therefore, the solution to the exponential equation 32x= 27 is x= 1.5.
Question 5
Question
Solve the exponential equation 23x+4 = 32 for x.
Solution
Step 1: Rewrite both sides of the equation with the same base.
23x+4 = 32
23x+4 = 25
Step 2: Set the exponents equal to each other.
3x+ 4 = 5
Step 3: Solve for x.
3x+ 4 = 5
3x= 1
x=1
3
Therefore, the solution to the equation 23x+4 = 32 is x=1
3.
Question 6
Question
Suppose a population of bacteria doubles every hour. If there are initially 100
bacteria, how many bacteria will there be after 6 hours?
4
Solution
Step 1: To model the population growth, we can use the exponential function
P(t) = P0·2t, where P(t)is the population at time t,P0is the initial population
(100 bacteria in this case), and tis the time in hours.
Step 2: Substituting P0= 100 and t= 6 into the exponential function, we
get:
P(6) = 100 ·26
Step 3: Calculating the expression gives:
P(6) = 100 ·64 = 6400
Step 4: Therefore, after 6 hours, there will be 6400 bacteria in the popula-
tion.
Question 7
Question
Solve the equation 4x−1= 8 for x.
Solution
Step 1: Rewrite the equation in terms of the base 4.
4x−1= 8
4x−1= 43
2
Step 2: Since the bases are the same, we can set the exponents equal to each
other.
x−1 = 3
2
Step 3: Solve for x.
x=3
2+ 1
Step 4: Simplify.
x=3
2+2
2=5
2
Step 5: Therefore, the solution to the equation 4x−1= 8 is x=5
2.
Question 8
Question
Samantha invested $10,000 in a savings account that pays 3% interest com-
pounded quarterly. How much money will Samantha have in her account after
5 years?
5
Solution
Step 1: First, we need to determine the variables involved in the compound
interest formula. The formula for compound interest is given by:
A=P1 + r
nnt
where: - Ais the amount of money accumulated after tyears, including in-
terest, - Pis the principal amount (the initial amount of money), - ris the
annual interest rate (in decimal form), - nis the number of times that interest
is compounded per year, and - tis the time the money is invested for in years.
In this case, we have: - P= $10,000 (the initial investment), - r= 0.03 (3%
interest rate in decimal form), - n= 4 (quarterly compounding), and - t= 5 (5
years).
Step 2: Substitute the given values into the compound interest formula and
solve for A.
A= 10000 1 + 0.03
44∗5
Step 3: Calculate the amount in Samantha’s account after 5 years.
A= 10000 1 + 0.03
420
A= 10000 (1 + 0.0075)20
A= 10000 ×1.007520
A= 10000 ×1.161067
Therefore, Samantha will have approximately $11,610.67 in her account after
5 years.
Question 9
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite the equation in terms of the same base:
32x−1= 33
Step 2: Set the exponents equal to each other:
2x−1 = 3
6
Step 3: Solve for x:
2x= 4
x= 2
Step 4: Check the solution in the original equation:
32(2)−1= 33
34−1= 33
33= 33
27 = 27
Therefore, the solution to the equation is x= 2.
Question 10
Question
Suppose a bacteria culture has an initial population of 1000 and the population
doubles every 6 hours. Write an exponential function to model the population
of the bacteria culture after thours.
Solution
Step 1: Let P(t)represent the population of the bacteria culture after thours.
Since the population doubles every 6 hours, we have the growth factor r= 2.
Step 2: To find the initial population P0, we use the given information that
the initial population is 1000. Therefore, P0= 1000.
Step 3: The exponential function that models the population growth can be
written as:
P(t) = P0×(r)t
time for doubling
Step 4: Substituting P0= 1000,r= 2, and time for doubling = 6 hours into
the equation, we have:
P(t) = 1000 ×2t
6
Therefore, the exponential function that models the population of the bac-
teria culture after thours is P(t) = 1000 ×2t
6.
Question 11
Question
The population of a city is initially 500,000 people, and it is projected to grow
exponentially at a rate of 3% per year. Find a function that models the popu-
lation P(t)after tyears.
7
Solution
Step 1: Recall that the formula for exponential growth is given by P(t) =
P0·(1 + r)t, where: - P(t)is the population after tyears, - P0is the initial
population, - ris the growth rate as a decimal, and - tis the number of years.
Step 2: Substitute the given values into the formula: P(t) = 500,000 ·(1 +
0.03)t
Step 3: Simplify the expression: P(t) = 500,000 ·(1.03)t
Therefore, the function that models the population P(t)after tyears is given
by P(t) = 500,000 ·(1.03)t.
Question 12
Question
Solve the following exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 in terms of a base of 3.
27 = 33
Step 2: Substitute the new representation of 27 into the 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 x.
2x−1 = 3
2x= 4
x= 2
Step 5: Check the solution by substituting x= 2 back into the original
equation.
32(2)−1= 27
33= 27
27 = 27
Therefore, the solution to the exponential equation is x= 2.
8
Question 13
Question
Consider the function f(x) = 2x−3. Find the inverse function f−1(x)and
evaluate f−1(5).
Solution
Step 1: To find the inverse function f−1(x), we start by swapping xand yand
then solve for y.
x= 2y−3
Step 2: Add 3 to both sides to isolate 2y.
x+ 3 = 2y
Step 3: Take the logarithm of both sides to solve for y.
log(x+ 3) = log(2y)
log(x+ 3) = ylog(2)
y=log(x+ 3)
log(2)
Step 4: Therefore, the inverse function f−1(x)is:
f−1(x) = log(x+ 3)
log(2)
Step 5: To evaluate f−1(5), substitute x= 5 into the inverse function.
f−1(5) = log(5 + 3)
log(2)
f−1(5) = log(8)
log(2)
f−1(5) = 3 log(2)
log(2)
f−1(5) = 3
Therefore, f−1(5) = 3.
Question 14
Question
Samantha invested $10,000 in a savings account that earns 4% interest com-
pounded quarterly. How long will it take for her money to double?
9
Solution
Step 1: The formula for compound interest is given by:
A=P1 + r
nnt
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: In this case, Samantha’s initial investment is $10,000, the annual
interest rate is 4%, and interest is compounded quarterly (so n= 4). Therefore,
we have:
2·10,000 = 10,000 1 + 0.04
44t
Step 3: Simplify the equation:
20,000 = 10,000 (1 + 0.01)4t
2 = (1.01)4t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln (1.01)4t
ln(2) = 4tln(1.01)
Step 5: Solve for t:
t=ln(2)
4 ln(1.01)
Step 6: Calculate the value of tusing a calculator:
t≈0.6931
4·0.0099
t≈0.6931
0.0396
t≈17.5
Step 7: It will take approximately 17.5 years for Samantha’s money to double
in the savings account.
Question 15
Question
Consider the exponential function f(x) = 3 ·2x.
If the population of a town is modeled by this function, where xrepresents
the number of years since the year 2020, determine the population of the town
in the year 2050. Round your answer to the nearest whole number.
10
Solution
Step 1: To find the population of the town in the year 2050, we need to determine
the value of f(2050) by substituting x= 2050 −2020 = 30 into the function
f(x) = 3 ·2x.
Step 2: Substitute x= 30 into the function f(x):
f(30) = 3 ·230
Step 3: Calculate 230:
230 = 1073741824
Step 4: Substitute 230 = 1073741824 into the expression for f(30):
f(30) = 3 ·1073741824
Step 5: Calculate the population of the town in the year 2050 by multiplying
3 by 1073741824:
f(30) = 3221225472
Therefore, the population of the town in the year 2050 is approximately
3,221,225,472.
Question 16
Question
Solve the exponential equation 32x+ 3x−10 = 0 for x.
Solution
Step 1: Let u= 3x. Then our equation becomes u2+u−10 = 0.
Step 2: Factor the quadratic equation u2+u−10 = 0 to get (u+2)(u−5) = 0.
Step 3: Set each factor equal to zero and solve for u:
u+ 2 = 0 or u−5 = 0
Step 4: Solve the first equation u+ 2 = 0 to get u=−2.
Step 5: Solve the second equation u−5 = 0 to get u= 5.
Step 6: Recall that u= 3x. Substitute these values back to solve for x:
3x=−2or 3x= 5
Step 7: Neither of these equations have a real solution as an exponential
function is always positive. Therefore, the original equation 32x+ 3x−10 = 0
has no real solution.
11
Question 17
Question
Samantha invests $5000 in an account that earns 3.5% interest compounded
continuously. How long will it take for her money to double?
Solution
Step 1: We will use the continuous compound interest formula A=P·ert,
where: - Ais the amount of money after a certain time - Pis the principal
amount (initial investment) - ris the interest rate - tis the time in years
Step 2: Since Samantha wants her money to double, the amount of money
after doubling will be 2P. So, we can set up the equation as:
2P=P·e0.035t
Step 3: Now, we can solve for t.
e0.035t= 2
0.035t= ln(2)
t=ln(2)
0.035
Step 4: Calculating the value of tusing a calculator:
t≈0.6931
0.035 ≈19.802
Step 5: It will take approximately 19.802 years for Samantha’s money to
double in the account.
Question 18
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,we have 32x+1 = 33.
Step 2: Set the exponents equal to each other.
Setting the exponents equal to each other gives us 2x+ 1 = 3.
12
Step 3: Solve for x.
2x+ 1 = 3
2x= 2
x= 1.
Step 4: Check the solution.
We substitute x= 1 back into the original equation to check for extraneous solutions.
32(1)+1 = 27
33= 27
27 = 27 ✓
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 19
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4%, compounded continuously. How long will it take for her investment to
double?
Solution
Step 1: We can model the amount of money in Samantha’s account after tyears
using the formula for continuous compounding:
A(t) = P·ert
where Pis the principal amount (initial investment), ris the annual interest
rate (in decimal form), and tis time in years.
Step 2: The amount we want to calculate is when her investment doubles,
so we want to find tsuch that A(t)=2P. Substituting in the given values, we
have:
2P=P·e0.04t
2 = e0.04t
Step 3: To solve for t, we need to isolate the variable in the exponential
equation. Taking the natural logarithm of both sides will help us do this:
ln 2 = ln e0.04t
ln 2 = 0.04tln e
13
ln 2 = 0.04t
Step 4: Now, we can solve for tby dividing both sides by 0.04:
t=ln 2
0.04 ≈0.6931
0.04 ≈17.33 years
Therefore, it will take approximately 17.33 years for Samantha’s investment
to double in value.
Question 20
Question
Solve for xin the equation 23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Since 8 = 23, we have:
23x−1= 23
Step 2: Set the exponents equal to each other. Since the bases are the same,
we can set the exponents equal to each other:
3x−1 = 3
Step 3: Solve for x. Add 1 to both sides of the equation:
3x= 4
Step 4: Divide by 3 to solve for x.
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
Question 21
Question
A certain radioactive substance has a half-life of 10 days. If there are initially
200 grams of the substance, find an exponential function that models the amount
of the substance remaining after tdays.
14
Solution
Step 1: The general formula for exponential decay is given by A(t) = A0·(0.5) t
h,
where: A(t)is the amount of substance remaining after tdays, A0is the initial
amount of substance, and his the half-life of the substance.
Step 2: Substituting A0= 200 grams and h= 10 days into the formula, we
get A(t) = 200 ·(0.5) t
10 .
Therefore, the exponential function that models the amount of the substance
remaining after tdays is A(t) = 200 ·(0.5) t
10 .
Question 22
Question
Solve the exponential equation: 3x−5·3x−1+ 6 = 0.
Solution
Step 1: Notice that the given equation can be written as a quadratic equation
in terms of 3x. Let y= 3x, then the equation becomes y2−5y+ 6 = 0.
Step 2: Now, we need to solve the quadratic equation y2−5y+ 6 = 0. The
factored form is (y−2)(y−3) = 0.
Step 3: Setting each factor to zero gives y−2 = 0 or y−3 = 0.
Step 4: Solve for yin each case. y−2 = 0 ⇒y= 2 and y−3 = 0 ⇒y= 3.
Step 5: Recall that y= 3x. So, 3x= 2 or 3x= 3.
Step 6: Solve for xin each case. For 3x= 2, we have x= log3(2). For
3x= 3, we have x= 1.
Therefore, the solutions to the original equation 3x−5·3x−1+ 6 = 0 are
x= log3(2) and x= 1.
Question 23
Question
Solve the exponential equation 32x+1 = 81.
Solution
Step 1: Rewrite 81 as a power of 3.
81 = 34
Step 2: Set the equation equal to the rewritten form of 81.
32x+1 = 34
15
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 4
Step 4: Solve for x.
2x= 4 −1
2x= 3
x=3
2
x= 1.5
Step 5: Check the solution.
32(1.5)+1 = 34
33+1 = 34
34= 34
Since both sides are equal, the solution x= 1.5is correct.
Question 24
Question
Samantha invested $10,000 in a savings account that earns 4% annual interest
compounded continuously. How much will the investment be worth after 10
years?
Solution
Step 1: We can use the formula for continuous compounding:
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: In this case, P= $10,000,r= 0.04, and t= 10 years. Substituting
these values into the formula, we get:
A= 10000 ·e0.04·10.
Step 3: Simplifying further, we get:
A= 10000 ·e0.4.
16
Step 4: We can use the approximate value of e≈2.718 to approximate the
final amount:
A≈10000 ·2.7180.4.
Step 5: Calculating the exponential term, we get:
A≈10000 ·1.4918.
Step 6: Finally, we find:
A≈$14,918.
Therefore, after 10 years, Samantha’s investment will be worth approxi-
mately $14,918.
Question 25
Question
Let f(x) = 4(3)x. Find the inverse function f−1(x).
Solution
Step 1: Replace f(x)with yto get y= 4(3)x.
Step 2: Switch xand yto get x= 4(3)y.
Step 3: Solve for yby isolating (3)y:
x= 4(3)y
x
4= (3)y
Step 4: Take the natural logarithm of both sides to bring down the exponent:
ln x
4= ln ((3)y)
ln x
4=yln(3)
Step 5: Divide by ln(3) to solve for y:
y=ln x
4
ln(3)
Step 6: Replace ywith f−1(x)to get the inverse function:
f−1(x) = ln x
4
ln(3)
17
Question 26
Question
Solve the exponential equation 2x+1 −4·2x+ 4 = 0.
Solution
Step 1: Let’s rewrite the equation in terms of a single base (2x) to simplify the
calculations.
2·2x−4·2x+ 4 = 0
Step 2: Combining like terms, we have
2x−4·2x+ 4 = 0
Step 3: Factor out 2xfrom the equation to get
2x(1 −4) + 4 = 0
Step 4: Simplify further to obtain
−2x+ 4 = 0
Step 5: Move 4 to the other side to isolate 2xto get
2x= 4
Step 6: Rewrite the equation using the same base to obtain
22= 2x
Step 7: Since the bases are the same, we can equate the exponents to find x
2 = x
Step 8: Therefore, the solution to the exponential equation 2x+1−4·2x+4 = 0
is x= 2.
Question 27
Question
Samantha invested $10,000 in a savings account that pays 3
18
Solution
Step 1: First, we need to determine the interest rate per compounding period.
Since the interest is compounded monthly at a rate of 3
Step 2: Next, we use the formula for compound interest:
A=P1 + r
nnt
where: - Ais the amount of money accumulated after tyears, - Pis the
principal amount (initial investment), - ris the annual interest rate (as a deci-
mal), - nis the number of times that interest is compounded per year, and - t
is the time the money is invested for in years.
Step 3: Substitute the given values into the formula:
A= 10000 1 + 0.03
12 12·5
Step 4: Simplify the expression inside the parentheses:
A= 10000 (1 + 0.0025)60
Step 5: Calculate the value inside the parentheses:
A= 10000(1.0025)60
Step 6: Finally, compute the final amount:
A≈10000 ×1.1611778 ≈$11,611.78
Therefore, after 5 years, Samantha will have approximately $11,611.78 in
the savings account.
Question 28
Question
Solve the following exponential equation for x:52x−1= 25.
Solution
Step 1: Rewrite 25 as a power of 5.
25 = 52
Step 2: Substitute 52into the equation and solve for x.
52x−1= 52
19
Step 3: Since the bases are the same, set the exponents equal to each other.
2x−1 = 2
Step 4: Solve the equation for x.
2x−1 = 2
2x= 3
x=3
2
Step 5: Check the solution by substituting x=3
2back into the original
equation.
52( 3
2)−1= 52
53−1= 52
52= 52
Step 6: Since the left side equals the right side, the solution is x=3
2.
Question 29
Question
Solve the exponential equation 3x+2 = 27.
Solution
Step 1: Rewrite 27 as a power of 3. Step 2: Solve for x.
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33, we can rewrite the exponential equation as
3x+2 = 33
.
Step 2: Solve for x.
Since both sides have the same base (3), we can equate the exponents:
x+ 2 = 3.
Now, solve for x:
x= 3 −2
x= 1.
Therefore, the solution to the exponential equation 3x+2 = 27 is x= 1.
20
Question 30
Question
Suppose the population of a small town is modeled by the function P(t) = 5000·
1.02t, where trepresents the number of years since the start of the observation.
Find the rate at which the population is growing after 5 years.
Solution
Step 1: Calculate the derivative of the population function P(t)with respect to
time t.d
dt P(t) = d
dt (5000 ·1.02t)
Step 2: Use the exponential function rule d
dt (at) = at·ln(a)and apply it to
the function 1.02t.d
dt P(t) = 5000 ·1.02t·ln(1.02)
Step 3: Evaluate the rate at which the population is growing after 5 years
by substituting t= 5 into the expression we found in Step 2.
d
dt P(t)
t=5
= 5000 ·1.025·ln(1.02)
Step 4: Perform the calculations to find the rate of population growth after
5 years.
d
dt P(t)
t=5
= 5000 ·1.10408 ≈5520.4
Therefore, the rate at which the population is growing after 5 years is ap-
proximately 5520.4 individuals per year.
Question 31
Question
An investment of 5000ismadeinanaccountthatpays4
Solution
Let A(t)be the amount of money in the account after tyears. The formula for
exponential growth with continuous compounding is given by:
A(t) = P ert,
where: - Pis the initial investment ($5000), - ris the interest rate (4- tis the
time in years, and - A(t)is the amount after tyears.
21
We are looking for the time tit takes for the investment to double, so we
want A(t)to be twice the initial investment amount. Thus, we want to solve
the equation 2P=P ert for t.
Step 1: Plug in the given values and set up the equation.
2(5000) = 5000e0.04t
Step 2: Simplify the equation.
10000 = 5000e0.04t
Step 3: Divide by 5000 to isolate the exponential term.
2 = e0.04t
Step 4: Take the natural logarithm of both sides to solve for t.
ln(2) = lne0.04t
Step 5: Use the property ln(ex) = x.
ln(2) = 0.04t
Step 6: Solve for t.
t=ln(2)
0.04
Step 7: Calculate the approximate value of t.
t≈0.6931
0.04 ≈17.33 years
So, it will take approximately 17.33 years for the investment to double in
value.
Question 32
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the equation.
32x−1= 33
22
Step 3: Since the bases are equal, we can equate the exponents.
2x−1 = 3
Step 4: Solve for x.
2x= 4
x= 2
Step 5: Check the solution.
32(2)−1= 33
33= 27
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 33
Question
Solve the exponential equation 32x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,we have 32x= 33.
Step 2: Equate the exponents.
Setting the exponents equal to each other, we get 2x= 3.
Step 3: Solve for x.
Dividing both sides by 2, we find x=3
2.
Therefore, the solution to the exponential equation 32x= 27 is x=3
2.
Question 34
Question
Suppose a population of bacteria doubles every 6 hours. If the initial population
is 1000 bacteria, find an exponential model for the population P(t)in terms of
time tin hours.
23
Solution
Step 1: Determine the growth factor
Since the population doubles every 6 hours, the growth factor ris 2.
Step 2: Write the exponential model
The exponential model for population growth is given by the formula:
P(t) = P0·(1 + r)t
k
where - P(t)is the population at time t, - P0is the initial population, - ris the
growth factor, and - kis the time it takes for the population to double.
Plugging in the values we know: - P0= 1000 (initial population) - r= 2
(growth factor) - k= 6 (time for population to double)
We have:
P(t) = 1000 ·(1 + 2) t
6
Step 3: Simplify the model
Simplify the expression to get the exponential model:
P(t) = 1000 ·2t
6
Therefore, the exponential model for the population P(t)in terms of time t
is P(t) = 1000 ·2t
6.
Question 35
Question
Let y= 23x−2x. Find the x-coordinate of the vertex for the graph of the
function y.
Solution
Step 1: Rewrite the function yin terms of a single exponential term. Step 2:
Identify the general form of an exponential function y=ax. Step 3: Find the
x-coordinate of the vertex using the formula x=−b
2a.
Step 1: Rewrite the function yin terms of a single exponential term.
Given y= 23x−2x, we can rewrite it as y= (2x)3−2x.
Step 2: Identify the general form of an exponential function y=ax.
We can rewrite the function in the form y=axby letting a= 2x. Thus,
y=a3−a.
Step 3: Find the x-coordinate of the vertex using the formula x=−b
2a.
The x-coordinate of the vertex of the function y=a3−ais given by x=
−−1
2(1) =1
2.
Therefore, the x-coordinate of the vertex for the graph of the function yis
1
2.
24
