MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 5
Liberty University
Question 1
Question
Let f(x) = 2xand g(x) = log2(x). Find the value of (f◦g)(4).
Solution
Step 1: Start by finding g(4).
g(4) = log2(4) = 2
Step 2: Next, find f(2) using f(x) = 2x.
f(2) = 22= 4
Step 3: Finally, calculate (f◦g)(4) by plugging in the result from step 1 into
function f.
(f◦g)(4) = f(g(4)) = f(2) = 4
Therefore, (f◦g)(4) = 4.
Question 2
Question
The population of a city is modeled by the exponential function P(t) = 40,000 ·
e0.02t, where trepresents the number of years since 2010. Determine the popu-
lation of the city in the year 2030.
Solution
Step 1: We are given the exponential function for the population of the city:
P(t) = 40,000 ·e0.02t. We want to find the population in the year 2030, which
is 20 years after 2010. So, we need to find P(20). Step 2: Substitute t= 20 into
the function P(t)to find the population in the year 2030.
P(20) = 40,000 ·e0.02·20
Step 3: Simplify the expression.
P(20) = 40,000 ·e0.4
Step 4: Use the fact that e≈2.71828 to evaluate e0.4.
e0.4≈2.718280.4≈1.49182
Step 5: Substitute back the approximate value of e0.4into P(20).
P(20) ≈40,000 ·1.49182
Step 6: Calculate the population in the year 2030.
P(20) ≈59,672.8
Therefore, the population of the city in the year 2030 is approximately
59,673.
Question 3
Question
Solve the exponential equation: 23x−1= 16.
Solution
Step 1: Rewrite the equation in terms of the base 2.
23x−1= 16
23x−1= 24
Step 2: Set the exponents equal to each other.
3x−1 = 4
Step 3: Solve for x.
3x= 4 + 1
3x= 5
x=5
3
Therefore, the solution to the exponential equation 23x−1= 16 is x=5
3.
2
Question 4
Question
Samantha invests $10,000 in a savings account with an annual interest rate
of 5%, compounded continuously. How long will it take for her investment to
double in value?
Solution
Step 1: First, we need to determine the continuous compound interest formula.
The formula for the amount of money after tyears in an account with principal
P, annual interest rate r, and compounded continuously is given by:
A=P·ert
where Ais the amount of money in the account after tyears, Pis the principal
amount, ris the annual interest rate, and eis the base of the natural logarithm
system.
Step 2: Since Samantha wants her investment to double, the amount of
money in her account after doubling will be 2P, where Pis the initial principal
amount. Therefore, we can write:
2P=P·e0.05t
Step 3: Now, we can solve for tby dividing both sides of the equation by P
and then taking the natural logarithm of both sides to solve for t:
ln(2) = 0.05t
Step 4: Next, we isolate tby dividing both sides by 0.05:
ln(2)
0.05 =t
Step 5: Now, we can calculate the value of t:
t≈ln(2)
0.05
Step 6: Using a calculator, we find:
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 in value when the interest is compounded continuously.
3
Question 5
Question
Let f(x)=3xand g(x) = log5(x). Determine the value of xthat satisfies the
equation f(g(x)) = 81.
Solution
Step 1: Start by substituting g(x)into f(x)and set the expression equal to 81.
f(g(x)) = f(log5(x))
= 3log5(x)
=x
Step 2: Set x= 81 and solve for x.
x= 81
log5(x) = log5(81)
log5(x) = log5(54)
log5(x) = 4
x= 54
x= 625
Therefore, the value of xthat satisfies the equation f(g(x)) = 81 is x= 625.
Question 6
Question
Suppose a certain investment grows according to the formula A(t) = 5000·e0.08t,
where A(t)represents the amount of money in the investment after tyears.
a) Calculate the amount of money in the investment after 5 years. Round
your answer to the nearest dollar.
b) Determine how many years it will take for the investment to double in
value.
Solution
a) To find the amount of money in the investment after 5 years, we evaluate
A(5) in the given formula.
A(5) = 5000 ·e0.08·5
Step 1: Calculate 0.08 ·5 = 0.4.
A(5) = 5000 ·e0.4
4
Step 2: Evaluate e0.4.
A(5) = 5000 ·e0.4
= 5000 ·1.491824693
≈7459.12
Therefore, the amount of money in the investment after 5 years is approxi-
mately $7459.
b) To determine how many years it will take for the investment to double in
value, we need to find the value of twhen A(t) = 2 ·A(0) = 2 ·5000 = 10000.
2·A(0) = 5000 ·e0.08t
10000 = 5000 ·e0.08t
2 = e0.08t
Step 1: Take the natural logarithm on both sides to solve for t.
ln(2) = ln(e0.08t)
ln(2) = 0.08t
t=ln(2)
0.08
Step 2: Calculate t.
t=ln(2)
0.08
≈0.693147181
0.08
≈8.664
Therefore, it will take approximately 8.664 years for the investment to double
in value.
Question 7
Question
Samantha invests $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much money will be in the account after 10 years?
Solution
Step 1: First, we need to determine the formula for the amount of money in the
account after tyears when interest is compounded continuously. This formula
is given by:
A=P·ert
5
where: - Ais the final amount in the account, - Pis the principal amount (initial
investment), - ris the interest rate, - tis the time in years, - eis Euler’s number
(≈2.71828).
Step 2: Substituting P= 5000,r= 0.035 (since 3.5% is equivalent to 0.035
as a decimal), and t= 10 into the formula, we get:
A= 5000 ·e0.035·10
Step 3: Calculate the exponent 0.035 ·10 = 0.35:
A= 5000 ·e0.35
Step 4: Using the approximation e0.35 ≈1.419067:
A≈5000 ·1.419067
Step 5: Calculate the final amount in the account after 10 years:
A≈7095.335
Therefore, there will be approximately $7095.34 in the account after 10 years
when interest is compounded continuously.
Question 8
Question
Let f(x) = 3xand g(x) = log2(x). Find the value of f(g(8)).
Solution
Step 1: First, we need to find g(8) by plugging x= 8 into g(x).
Step 1: g(8) = log2(8)
Step 2: Since 23= 8, we can rewrite g(8) as:
Step 2: g(8) = log2(8) = log2(23) = 3
Step 3: Now, we need to find f(g(8)) by plugging x= 3 into f(x).
Step 3: f(g(8)) = f(3) = 33
Step 4: Finally, calculate 33.
Step 4: 33= 27
Therefore, the value of f(g(8)) is 27 .
6
Question 9
Question
Samantha has invested $10,000 in a savings account that offers 4% interest
compounded continuously. How much will she have in the account after 5 years?
Solution
Step 1: Use the formula for continuously compounded interest:
A=P ert
where: - Ais the amount of money accumulated after tyears, - Pis the prin-
cipal amount (initial investment), - ris the annual interest rate (decimal), -
tis the time the money is invested for (in years), and - eis Euler’s number
(approximately equal to 2.71828).
Step 2: Plug in the given values. Given: - P= $10,000, - r= 0.04 (4% as a
decimal), - t= 5 years.
We have:
A= 10000 ×e0.04×5
Step 3: Calculate the exponent first:
0.04 ×5 = 0.2
Step 4: Plug this back into the formula and solve for A:
A= 10000 ×e0.2
Step 5: Calculate e0.2≈1.221402 (rounding to six decimal places).
Step 6: Multiply to find the final amount:
A≈10000 ×1.221402 ≈$12,214.02
Therefore, Samantha will have approximately $12,214.02 in the account after
5 years.
Question 10
Question
A certain radioactive substance decays according to the formula A(t) = A0e−kt,
where A(t)represents the amount of substance at time t,A0is the initial
amount, and kis a positive constant. If 50 grams of a substance decays to
40 grams after 2 days, find the value of kfor this substance.
7
Solution
Step 1: We are given that the initial amount of substance is A0= 50 grams,
the amount after 2 days is A(2) = 40 grams, and the time is t= 2 days. We
can plug these values into the radioactive decay formula to get an equation to
solve for k:
40 = 50e−2k
Step 2: Divide both sides of the equation by 50 to isolate the exponential
term:
0.8 = e−2k
Step 3: Take the natural logarithm of both sides to solve for k:
ln(0.8) = ln(e−2k)
Step 4: Use the property of logarithms ln(ex) = xto simplify the equation:
ln(0.8) = −2k
Step 5: Solve for kby dividing both sides by −2:
k=ln(0.8)
−2
Step 6: Use a calculator to approximate the value of k:
k≈ln(0.8)
−2≈−0.2231
−2≈0.1115
Therefore, the value of kfor this radioactive substance is approximately
0.1115.
Question 11
Question
Solve the exponential equation 32x−3 = 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute this into the original equation:
32x−3 = 33
Step 3: Use the properties of exponents to simplify the equation:
32x= 33+ 3
8
32x= 34
Step 4: Set the exponents equal to each other:
2x= 4
Step 5: Solve for x:
x=4
2
x= 2
Therefore, the solution to the equation 32x−3 = 27 is x= 2.
Question 12
Question
Solve the exponential equation 32x= 27.
Solution
Step 1: Express 27 as a power of 3.
27 = 33
Step 2: Set the exponent 2xequal to 3.
2x= 3
Step 3: Solve for xby dividing both sides by 2.
x=3
2
Therefore, the solution to the exponential equation 32x= 27 is x=3
2.
Question 13
Question
Let f(x) = 3xand g(x) = log3(x). Determine the domain of f(g(x)).
9
Solution
Step 1: To find the domain of f(g(x)), we need to consider the restrictions on
the composition function.
Step 2: Since g(x) = log3(x), we know that g(x)is defined only for positive
values of x, i.e., x > 0.
Step 3: Now, we need to substitute g(x)into f(x)to find f(g(x)).
f(g(x)) = f(log3(x)) = 3log3(x)
Step 4: Remember that 3log3(x)=x.
Step 5: Therefore, the domain of f(g(x)) is all positive real numbers. In
interval notation, the domain is (0,∞).
Question 14
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where t
represents the number of years since the population started being tracked. Find
the population of the city after 10 years.
Solution
Step 1: Substitute t= 10 into the formula P(t) = 5000 ·1.02tto find the
population after 10 years.
P(10) = 5000 ·1.0210
= 5000 ·1.218994
= 6094.97
Step 2: Round the population to the nearest whole number since a popula-
tion count must be a whole number. Therefore, the population of the city after
10 years is approximately 6095.
Question 15
Question
Solve the exponential equation 2x−3·2x−1−4 = 0.
Solution
Step 1: Let’s rewrite the equation by factoring out 2x−1:
2x−3·2x−1−4 = 0
10
2x−1(2 −3) −4 = 0
2x−1−4 = 0
Step 2: Now, we have a simple exponential equation. Let’s solve for 2x−1:
2x−1= 4
Step 3: Rewrite 4as 22:
2x−1= 22
Step 4: Since the bases are the same, we can equate the exponents:
x−1 = 2
Step 5: Solve for x:
x= 2 + 1
x= 3
Step 6: Therefore, the solution to the given exponential equation is x= 3.
Question 16
Question
Suppose an investment of $5000 is made in an account that pays 3% interest,
compounded continuously. How long will it take for the investment to double
in value?
Solution
Let’s denote the initial investment amount as P= $5000, the annual interest
rate as r= 0.03 (3%), and the time required to double the investment as t.
We use the continuous compounding formula for exponential growth:
A=P ert
where Ais the final amount, Pis the principal amount, eis the base of the
natural logarithm, ris the interest rate, and tis time in years.
Step 1: We want the investment to double, so the final amount will be
2P= 2 ×5000 = $10000. Therefore, our equation becomes:
10000 = 5000e0.03t
Step 2: Divide both sides of the equation by 5000 to isolate the exponential
term:
2 = e0.03t
11
Step 3: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.03t)
ln(2) = 0.03t
t=ln(2)
0.03
Step 4: Now, calculate the value of t:
t=ln(2)
0.03 ≈0.6931
0.03 ≈23.1045
Therefore, it will take approximately 23.1 years for the investment to double
in value when continuously compounded at 3% interest.
Question 17
Question
Solve the exponential equation 3x−4= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation:
3x−4= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
x−4 = 3
Step 4: Solve for x:
x= 3 + 4
x= 7
Step 5: Therefore, the solution to the exponential equation 3x−4= 27 is
x= 7.
Question 18
Question
Solve for x:e2x−5ex+ 6 = 0.
12
Solution
Step 1: Let u=ex. Then, the equation becomes u2−5u+ 6 = 0.
Step 2: Factor the quadratic equation: (u−2)(u−3) = 0.
Step 3: Set each factor to zero: u−2 = 0 or u−3 = 0.
Step 4: Solve for uin each case: For u−2 = 0:u= 2. For u−3 = 0:u= 3.
Step 5: Recall that u=ex.
Step 6: Solve for x: For u= 2:ex= 2 =⇒x= ln(2). For u= 3:
ex= 3 =⇒x= ln(3).
Therefore, the solutions to the equation e2x−5ex+ 6 = 0 are x= ln(2) and
x= ln(3).
Question 19
Question
Suppose an investment of $5000 grows exponentially with a continuous interest
rate of 6% per year. How much will the investment be worth after 10 years?
Solution
Step 1: Recall the formula for compound interest in continuous compounding:
A=P·ert
where: A= the amount of money accumulated after tyears, including interest
P= the principal amount (initial investment) r= the annual interest rate
(expressed as a decimal) t= the time the money is invested for
Step 2: Substitute the given values into the formula: The principal amount,
P= $5000 The annual interest rate, r= 0.06 The time period, t= 10 years
Step 3: Plug the values into the formula and calculate:
A= 5000 ·e0.06·10
A= 5000 ·e0.6
A≈5000 ·1.82212
A≈9110.6
Therefore, the investment will be worth approximately $9110.6after 10
years.
Question 20
Question
The population of a city can be modeled by the function P(t) = 2500 ·1.03t,
where trepresents the number of years since the year 2020. At what rate is the
population growing in 2025?
13
Solution
Step 1: Calculate the population in 2025 using the given function.
Step 2: Find the population in 2026 by plugging in t= 6 in the function and
calculate the difference between the populations for 2025 and 2026.
Step 3: Calculate the rate of population growth by taking the difference between
the populations and dividing it by 1 year.
Step 1: Calculate the population in 2025:
P(5) = 2500 ·1.035= 2500 ·1.159274 = 2898.185.
The population in 2025 is approximately 2898.185.
Step 2: Calculate the population in 2026:
P(6) = 2500 ·1.036= 2500 ·1.19101622 = 2977.54.
The population in 2026 is approximately 2977.54.
The rate of population growth between 2025 and 2026 is:
2977.54 −2898.185 = 79.355.
Step 3: Calculate the rate of population growth:
79.355
1= 79.355.
Therefore, the rate at which the population is growing in 2025 is approxi-
mately 79.355 people per year.
Question 21
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: Set the equation equal to the new representation of 27.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
14
Step 4: Solve for xby isolating the variable.
2x+ 1 = 3
2x= 2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation.
32(1)+1 = 33
33= 27
Therefore, the solution to the equation is x= 1.
Question 22
Question
Solve the exponential equation 3x−1+ 3x−2= 10.
Solution
Step 1: Let’s rewrite the equation in terms of a single base:
3x−1+ 3x−2= 3 ·3x−2+ 3x−2= 3 ·3x−2+ 3x−2= 3x−2(3 + 1) = 4 ·3x−2
Step 2: Rewrite the equation as 4·3x−2= 10.
Step 3: Divide by 4 to isolate 3x−2:
3x−2=10
4=5
2
Step 4: Write the right-hand side as 31−log32:
Step 5: Since both sides have the same base, we can set the exponents equal
to each other:
x−2 = 1 −log32
Step 6: Add 2 to both sides to solve for x:
x= 1 + 2 −log32 = 3 −log32
Hence, the solution to the exponential equation 3x−1+ 3x−2= 10 is x=
3−log32.
Question 23
Question
Samantha invested $10,000 in an account that earns 3% compounded continu-
ously. How much will the investment be worth after 5 years?
15
Solution
Step 1: Identify the formula for continuous compounding which is given by:
A=P·ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= the annual interest rate (in decimal form), and
t= the time the money is invested for.
Step 2: Plug in the given values into the formula: P= $10,000,r= 0.03
(3% as a decimal), and t= 5 years.
Step 3: Calculate the amount accumulated after 5 years using the formula.
A= $10,000 ·e0.03·5
Step 4: Simplify the expression and calculate the final amount.
A= $10,000 ·e0.15
A≈$10,000 ·1.161834
A≈$11,618.34
Therefore, the investment will be worth approximately $11,618.34 after 5
years when compounded continuously.
Question 24
Question
Solve the exponential equation for x:4(5x)−20 = 5x+1.
Solution
Step 1: Let’s rewrite the given equation by recognizing that 5x+1 = 5 ·5x.
4·5x−20 = 5 ·5x
Step 2: Subtract 5xfrom both sides of the equation to isolate 5xon one side.
4·5x−5·5x= 20
Step 3: Combine like terms on the left side of the equation.
4·5x−5·5x= 20
−1·5x= 20
Step 4: Divide both sides by −1to solve for 5x.
5x=−20
Step 5: Since the base of the exponential expression is 5, which is greater
than 1, there are no real solutions for x. Thus, the equation 4(5x)−20 = 5x+1
has no solution in the real number system.
16
Question 25
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33back into the equation and solve for x:
32x−1= 33
Step 3: By the property of equality for exponential functions, we can equate
the exponents and solve for x:
2x−1 = 3
Step 4: Add 1 to both sides of the equation:
2x= 4
Step 5: Divide by 2 to solve for x:
x= 2
Step 6: Therefore, the solution to the exponential equation 32x−1= 27 is
x= 2.
Question 26
Question
Solve the exponential equation 52x−1= 125.
Solution
Step 1: Rewrite the equation using the same base on both sides.
52x−1= 125
52x−1= 53
Step 2: Set the exponents equal to each other.
2x−1 = 3
17
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 4: Verify the solution. Substitute x= 2 back into the original equation:
52(2)−1= 125
54−1= 125
53= 125
125 = 125
Therefore, the solution to the exponential equation is x= 2.
Question 27
Question
Solve for xin the equation 32x= 27.
Solution
Step 1: Rewrite 27 using the base 3. Step 2: Use the properties of exponents
to solve for x.
Question 28
Question
Solve the following exponential equation for x:
23x−1= 8x+2
Solution
Step 1: Rewrite 8as a power of 2.
23x−1= (23)x+2
Step 2: Simplify the right side of the equation.
23x−1= 23(x+2)
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
3x−1 = 3(x+ 2)
18
Step 4: Expand and solve for x.
3x−1 = 3x+ 6
−1 = 6
Step 5: Since we reached a false statement, there is no solution to the equa-
tion.
Question 29
Question
Solve the exponential equation for x:32x+1 −10·3x+9 = 0. Round your answer
to three decimal places if necessary.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation to solve for
3x. Let y= 3x. Then, we have y2−10y+ 9 = 0.
Step 2: Now, let’s solve the quadratic equation y2−10y+ 9 = 0 using the
quadratic formula: y=−(−10)±√(−10)2−4(1)(9)
2(1) .
Step 3: Simplifying gives y=10±√100−36
2.
Step 4: Therefore, we have two possible values for y:y1=10+√64
2and
y2=10−√64
2.
Step 5: Solving for y1gives y1=10+8
2= 9, and solving for y2gives y2=
10−8
2= 1.
Step 6: Recall that y= 3x. Therefore, we have two possible exponential
equations to solve: 3x= 9 and 3x= 1.
Step 7: Solving 3x= 9 gives x= log3(9) = 2, and solving 3x= 1 gives
x= log3(1) = 0.
Step 8: Thus, the solutions to the original equation 32x+1 −10 ·3x+ 9 = 0
are x= 2 and x= 0.
Question 30
Question
Solve the following exponential equation for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2:
23x−1= 23
19
Step 2: Since the bases are the same, we can equate the exponents:
3x−1 = 3
Step 3: Solve for x:
3x−1 = 3
3x= 4
x=4
3
Step 4: Therefore, the solution to the given exponential equation is x=4
3.
Question 31
Question
Solve the exponential equation: 32x+1 = 9.
Solution
Step 1: Rewrite 9 as a power of 3.
9 = 32
Step 2: Substitute 32back into the equation.
32x+1 = 32
Step 3: Set the exponents equal to each other.
2x+ 1 = 2
Step 4: Solve for x.
2x= 2 −1
2x= 1
x=1
2
Step 5: Check the solution by substituting x=1
2back into the original
equation.
32( 1
2)+1 = 32
32= 9
Since the left side equals the right side, the solution x=1
2is correct.
20
Question 32
Question
Suppose the population of a city can be modeled by the function P(t) = 8000 ·
e0.02t, where tis the number of years since 2020. Find the population of the
city in the year 2030.
Solution
Step 1: To find the population of the city in the year 2030, we need to substitute
t= 10 into the population function P(t).
P(10) = 8000 ·e0.02·10
Step 2: Calculate the exponential term.
P(10) = 8000 ·e0.2
Step 3: Evaluate e0.2using a calculator or by approximating it as follows:
e0.2≈1.221
Step 4: Substitute the value of e0.2back into the equation.
P(10) = 8000 ·1.221
Step 5: Multiply to find the population.
P(10) = 9768
Therefore, the population of the city in the year 2030 is 9768.
Question 33
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where
tis the number of years since the year 2010.
a) Determine the population of the city in the year 2025.
b) In what year will the population reach 10,000?
Solution
a) To determine the population in the year 2025, we need to find P(15) since
2025 is 15 years after 2010.
P(15) = 5000 ·e0.02·15
21
Step 1: Calculate the exponent inside the function.
0.02 ·15 = 0.3
Step 2: Substitute the exponent back into the equation and simplify.
P(15) = 5000 ·e0.3
Step 3: Calculate e0.3as an approximation.
e0.3≈1.3499
Step 4: Substitute the approximation back into the equation and find the
population.
P(15) = 5000 ·1.3499 = 6749.5
Therefore, the population of the city in the year 2025 is approximately
6749.5.
b) To find the year when the population reaches 10,000, we need to solve
the equation 5000 ·e0.02t= 10000.
Step 1: Divide both sides by 5000 to isolate the exponential term.
e0.02t= 2
Step 2: Take the natural logarithm of both sides to remove the base e.
ln(e0.02t)= ln(2)
Step 3: Use the property of logarithms to bring down the exponent.
0.02t·ln(e) = ln(2)
Step 4: Since ln(e) = 1, divide both sides by 0.02 to solve for t.
t=ln(2)
0.02 ≈34.66
Therefore, the population will reach 10,000 in the year 2044 (rounded up).
Question 34
Question
Solve the exponential equation 4x−1= 8.
22
Solution
Step 1: Rewrite 8as a power of 4using the property 43= 64.
Step 2: Substitute 8as 42in the equation.
(42)x−1= 43
Step 3: Simplify the left side using the property (am)n=am·n.
42(x−1) = 43
Step 4: Apply the power rule for exponents by multiplying the exponents.
42x−2= 43
Step 5: Since the bases are the same, set the exponents equal to each other.
2x−2 = 3
Step 6: Solve for xby isolating the variable.
2x= 5
x=5
2
Step 7: Therefore, the solution to the exponential equation 4x−1= 8 is
x=5
2.
Question 35
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where t
represents the number of years since the population was recorded. What is the
population after 10 years to the nearest whole number?
Solution
Step 1: Substitute t= 10 into the population function to find P(10). Step 2:
Calculate the population after 10 years by simplifying the expression. Step 3:
Round the final answer to the nearest whole number.
Step 1: Substitute t= 10 into the population function:
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression:
P(10) = 5000 ·e0.2
Step 3: Calculate the value of P(10):
P(10) ≈5000 ·1.221402758 ≈6107
Therefore, the population of the city after 10 years is approximately 6107
people.
23
Question 4
Question
Samantha invests $10,000 in a savings account with an annual interest rate
of 5%, compounded continuously. How long will it take for her investment to
double in value?
Solution
Step 1: First, we need to determine the continuous compound interest formula.
The formula for the amount of money after tyears in an account with principal
P, annual interest rate r, and compounded continuously is given by:
A=P·ert
where Ais the amount of money in the account after tyears, Pis the principal
amount, ris the annual interest rate, and eis the base of the natural logarithm
system.
Step 2: Since Samantha wants her investment to double, the amount of
money in her account after doubling will be 2P, where Pis the initial principal
amount. Therefore, we can write:
2P=P·e0.05t
Step 3: Now, we can solve for tby dividing both sides of the equation by P
and then taking the natural logarithm of both sides to solve for t:
ln(2) = 0.05t
Step 4: Next, we isolate tby dividing both sides by 0.05:
ln(2)
0.05 =t
Step 5: Now, we can calculate the value of t:
t≈ln(2)
0.05
Step 6: Using a calculator, we find:
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 in value when the interest is compounded continuously.
3
Question 5
Question
Let f(x)=3xand g(x) = log5(x). Determine the value of xthat satisfies the
equation f(g(x)) = 81.
Solution
Step 1: Start by substituting g(x)into f(x)and set the expression equal to 81.
f(g(x)) = f(log5(x))
= 3log5(x)
=x
Step 2: Set x= 81 and solve for x.
x= 81
log5(x) = log5(81)
log5(x) = log5(54)
log5(x) = 4
x= 54
x= 625
Therefore, the value of xthat satisfies the equation f(g(x)) = 81 is x= 625.
Question 6
Question
Suppose a certain investment grows according to the formula A(t) = 5000·e0.08t,
where A(t)represents the amount of money in the investment after tyears.
a) Calculate the amount of money in the investment after 5 years. Round
your answer to the nearest dollar.
b) Determine how many years it will take for the investment to double in
value.
Solution
a) To find the amount of money in the investment after 5 years, we evaluate
A(5) in the given formula.
A(5) = 5000 ·e0.08·5
Step 1: Calculate 0.08 ·5 = 0.4.
A(5) = 5000 ·e0.4
4
Step 2: Evaluate e0.4.
A(5) = 5000 ·e0.4
= 5000 ·1.491824693
≈7459.12
Therefore, the amount of money in the investment after 5 years is approxi-
mately $7459.
b) To determine how many years it will take for the investment to double in
value, we need to find the value of twhen A(t) = 2 ·A(0) = 2 ·5000 = 10000.
2·A(0) = 5000 ·e0.08t
10000 = 5000 ·e0.08t
2 = e0.08t
Step 1: Take the natural logarithm on both sides to solve for t.
ln(2) = ln(e0.08t)
ln(2) = 0.08t
t=ln(2)
0.08
Step 2: Calculate t.
t=ln(2)
0.08
≈0.693147181
0.08
≈8.664
Therefore, it will take approximately 8.664 years for the investment to double
in value.
Question 7
Question
Samantha invests $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much money will be in the account after 10 years?
Solution
Step 1: First, we need to determine the formula for the amount of money in the
account after tyears when interest is compounded continuously. This formula
is given by:
A=P·ert
5
where: - Ais the final amount in the account, - Pis the principal amount (initial
investment), - ris the interest rate, - tis the time in years, - eis Euler’s number
(≈2.71828).
Step 2: Substituting P= 5000,r= 0.035 (since 3.5% is equivalent to 0.035
as a decimal), and t= 10 into the formula, we get:
A= 5000 ·e0.035·10
Step 3: Calculate the exponent 0.035 ·10 = 0.35:
A= 5000 ·e0.35
Step 4: Using the approximation e0.35 ≈1.419067:
A≈5000 ·1.419067
Step 5: Calculate the final amount in the account after 10 years:
A≈7095.335
Therefore, there will be approximately $7095.34 in the account after 10 years
when interest is compounded continuously.
Question 8
Question
Let f(x) = 3xand g(x) = log2(x). Find the value of f(g(8)).
Solution
Step 1: First, we need to find g(8) by plugging x= 8 into g(x).
Step 1: g(8) = log2(8)
Step 2: Since 23= 8, we can rewrite g(8) as:
Step 2: g(8) = log2(8) = log2(23) = 3
Step 3: Now, we need to find f(g(8)) by plugging x= 3 into f(x).
Step 3: f(g(8)) = f(3) = 33
Step 4: Finally, calculate 33.
Step 4: 33= 27
Therefore, the value of f(g(8)) is 27 .
6
Question 9
Question
Samantha has invested $10,000 in a savings account that offers 4% interest
compounded continuously. How much will she have in the account after 5 years?
Solution
Step 1: Use the formula for continuously compounded interest:
A=P ert
where: - Ais the amount of money accumulated after tyears, - Pis the prin-
cipal amount (initial investment), - ris the annual interest rate (decimal), -
tis the time the money is invested for (in years), and - eis Euler’s number
(approximately equal to 2.71828).
Step 2: Plug in the given values. Given: - P= $10,000, - r= 0.04 (4% as a
decimal), - t= 5 years.
We have:
A= 10000 ×e0.04×5
Step 3: Calculate the exponent first:
0.04 ×5 = 0.2
Step 4: Plug this back into the formula and solve for A:
A= 10000 ×e0.2
Step 5: Calculate e0.2≈1.221402 (rounding to six decimal places).
Step 6: Multiply to find the final amount:
A≈10000 ×1.221402 ≈$12,214.02
Therefore, Samantha will have approximately $12,214.02 in the account after
5 years.
Question 10
Question
A certain radioactive substance decays according to the formula A(t) = A0e−kt,
where A(t)represents the amount of substance at time t,A0is the initial
amount, and kis a positive constant. If 50 grams of a substance decays to
40 grams after 2 days, find the value of kfor this substance.
7
Solution
Step 1: We are given that the initial amount of substance is A0= 50 grams,
the amount after 2 days is A(2) = 40 grams, and the time is t= 2 days. We
can plug these values into the radioactive decay formula to get an equation to
solve for k:
40 = 50e−2k
Step 2: Divide both sides of the equation by 50 to isolate the exponential
term:
0.8 = e−2k
Step 3: Take the natural logarithm of both sides to solve for k:
ln(0.8) = ln(e−2k)
Step 4: Use the property of logarithms ln(ex) = xto simplify the equation:
ln(0.8) = −2k
Step 5: Solve for kby dividing both sides by −2:
k=ln(0.8)
−2
Step 6: Use a calculator to approximate the value of k:
k≈ln(0.8)
−2≈−0.2231
−2≈0.1115
Therefore, the value of kfor this radioactive substance is approximately
0.1115.
Question 11
Question
Solve the exponential equation 32x−3 = 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute this into the original equation:
32x−3 = 33
Step 3: Use the properties of exponents to simplify the equation:
32x= 33+ 3
8
32x= 34
Step 4: Set the exponents equal to each other:
2x= 4
Step 5: Solve for x:
x=4
2
x= 2
Therefore, the solution to the equation 32x−3 = 27 is x= 2.
Question 12
Question
Solve the exponential equation 32x= 27.
Solution
Step 1: Express 27 as a power of 3.
27 = 33
Step 2: Set the exponent 2xequal to 3.
2x= 3
Step 3: Solve for xby dividing both sides by 2.
x=3
2
Therefore, the solution to the exponential equation 32x= 27 is x=3
2.
Question 13
Question
Let f(x) = 3xand g(x) = log3(x). Determine the domain of f(g(x)).
9
Solution
Step 1: To find the domain of f(g(x)), we need to consider the restrictions on
the composition function.
Step 2: Since g(x) = log3(x), we know that g(x)is defined only for positive
values of x, i.e., x > 0.
Step 3: Now, we need to substitute g(x)into f(x)to find f(g(x)).
f(g(x)) = f(log3(x)) = 3log3(x)
Step 4: Remember that 3log3(x)=x.
Step 5: Therefore, the domain of f(g(x)) is all positive real numbers. In
interval notation, the domain is (0,∞).
Question 14
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where t
represents the number of years since the population started being tracked. Find
the population of the city after 10 years.
Solution
Step 1: Substitute t= 10 into the formula P(t) = 5000 ·1.02tto find the
population after 10 years.
P(10) = 5000 ·1.0210
= 5000 ·1.218994
= 6094.97
Step 2: Round the population to the nearest whole number since a popula-
tion count must be a whole number. Therefore, the population of the city after
10 years is approximately 6095.
Question 15
Question
Solve the exponential equation 2x−3·2x−1−4 = 0.
Solution
Step 1: Let’s rewrite the equation by factoring out 2x−1:
2x−3·2x−1−4 = 0
10
2x−1(2 −3) −4 = 0
2x−1−4 = 0
Step 2: Now, we have a simple exponential equation. Let’s solve for 2x−1:
2x−1= 4
Step 3: Rewrite 4as 22:
2x−1= 22
Step 4: Since the bases are the same, we can equate the exponents:
x−1 = 2
Step 5: Solve for x:
x= 2 + 1
x= 3
Step 6: Therefore, the solution to the given exponential equation is x= 3.
Question 16
Question
Suppose an investment of $5000 is made in an account that pays 3% interest,
compounded continuously. How long will it take for the investment to double
in value?
Solution
Let’s denote the initial investment amount as P= $5000, the annual interest
rate as r= 0.03 (3%), and the time required to double the investment as t.
We use the continuous compounding formula for exponential growth:
A=P ert
where Ais the final amount, Pis the principal amount, eis the base of the
natural logarithm, ris the interest rate, and tis time in years.
Step 1: We want the investment to double, so the final amount will be
2P= 2 ×5000 = $10000. Therefore, our equation becomes:
10000 = 5000e0.03t
Step 2: Divide both sides of the equation by 5000 to isolate the exponential
term:
2 = e0.03t
11
Step 3: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.03t)
ln(2) = 0.03t
t=ln(2)
0.03
Step 4: Now, calculate the value of t:
t=ln(2)
0.03 ≈0.6931
0.03 ≈23.1045
Therefore, it will take approximately 23.1 years for the investment to double
in value when continuously compounded at 3% interest.
Question 17
Question
Solve the exponential equation 3x−4= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation:
3x−4= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
x−4 = 3
Step 4: Solve for x:
x= 3 + 4
x= 7
Step 5: Therefore, the solution to the exponential equation 3x−4= 27 is
x= 7.
Question 18
Question
Solve for x:e2x−5ex+ 6 = 0.
12
Solution
Step 1: Let u=ex. Then, the equation becomes u2−5u+ 6 = 0.
Step 2: Factor the quadratic equation: (u−2)(u−3) = 0.
Step 3: Set each factor to zero: u−2 = 0 or u−3 = 0.
Step 4: Solve for uin each case: For u−2 = 0:u= 2. For u−3 = 0:u= 3.
Step 5: Recall that u=ex.
Step 6: Solve for x: For u= 2:ex= 2 =⇒x= ln(2). For u= 3:
ex= 3 =⇒x= ln(3).
Therefore, the solutions to the equation e2x−5ex+ 6 = 0 are x= ln(2) and
x= ln(3).
Question 19
Question
Suppose an investment of $5000 grows exponentially with a continuous interest
rate of 6% per year. How much will the investment be worth after 10 years?
Solution
Step 1: Recall the formula for compound interest in continuous compounding:
A=P·ert
where: A= the amount of money accumulated after tyears, including interest
P= the principal amount (initial investment) r= the annual interest rate
(expressed as a decimal) t= the time the money is invested for
Step 2: Substitute the given values into the formula: The principal amount,
P= $5000 The annual interest rate, r= 0.06 The time period, t= 10 years
Step 3: Plug the values into the formula and calculate:
A= 5000 ·e0.06·10
A= 5000 ·e0.6
A≈5000 ·1.82212
A≈9110.6
Therefore, the investment will be worth approximately $9110.6after 10
years.
Question 20
Question
The population of a city can be modeled by the function P(t) = 2500 ·1.03t,
where trepresents the number of years since the year 2020. At what rate is the
population growing in 2025?
13
Solution
Step 1: Calculate the population in 2025 using the given function.
Step 2: Find the population in 2026 by plugging in t= 6 in the function and
calculate the difference between the populations for 2025 and 2026.
Step 3: Calculate the rate of population growth by taking the difference between
the populations and dividing it by 1 year.
Step 1: Calculate the population in 2025:
P(5) = 2500 ·1.035= 2500 ·1.159274 = 2898.185.
The population in 2025 is approximately 2898.185.
Step 2: Calculate the population in 2026:
P(6) = 2500 ·1.036= 2500 ·1.19101622 = 2977.54.
The population in 2026 is approximately 2977.54.
The rate of population growth between 2025 and 2026 is:
2977.54 −2898.185 = 79.355.
Step 3: Calculate the rate of population growth:
79.355
1= 79.355.
Therefore, the rate at which the population is growing in 2025 is approxi-
mately 79.355 people per year.
Question 21
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: Set the equation equal to the new representation of 27.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
14
Step 4: Solve for xby isolating the variable.
2x+ 1 = 3
2x= 2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation.
32(1)+1 = 33
33= 27
Therefore, the solution to the equation is x= 1.
Question 22
Question
Solve the exponential equation 3x−1+ 3x−2= 10.
Solution
Step 1: Let’s rewrite the equation in terms of a single base:
3x−1+ 3x−2= 3 ·3x−2+ 3x−2= 3 ·3x−2+ 3x−2= 3x−2(3 + 1) = 4 ·3x−2
Step 2: Rewrite the equation as 4·3x−2= 10.
Step 3: Divide by 4 to isolate 3x−2:
3x−2=10
4=5
2
Step 4: Write the right-hand side as 31−log32:
Step 5: Since both sides have the same base, we can set the exponents equal
to each other:
x−2 = 1 −log32
Step 6: Add 2 to both sides to solve for x:
x= 1 + 2 −log32 = 3 −log32
Hence, the solution to the exponential equation 3x−1+ 3x−2= 10 is x=
3−log32.
Question 23
Question
Samantha invested $10,000 in an account that earns 3% compounded continu-
ously. How much will the investment be worth after 5 years?
15
Solution
Step 1: Identify the formula for continuous compounding which is given by:
A=P·ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= the annual interest rate (in decimal form), and
t= the time the money is invested for.
Step 2: Plug in the given values into the formula: P= $10,000,r= 0.03
(3% as a decimal), and t= 5 years.
Step 3: Calculate the amount accumulated after 5 years using the formula.
A= $10,000 ·e0.03·5
Step 4: Simplify the expression and calculate the final amount.
A= $10,000 ·e0.15
A≈$10,000 ·1.161834
A≈$11,618.34
Therefore, the investment will be worth approximately $11,618.34 after 5
years when compounded continuously.
Question 24
Question
Solve the exponential equation for x:4(5x)−20 = 5x+1.
Solution
Step 1: Let’s rewrite the given equation by recognizing that 5x+1 = 5 ·5x.
4·5x−20 = 5 ·5x
Step 2: Subtract 5xfrom both sides of the equation to isolate 5xon one side.
4·5x−5·5x= 20
Step 3: Combine like terms on the left side of the equation.
4·5x−5·5x= 20
−1·5x= 20
Step 4: Divide both sides by −1to solve for 5x.
5x=−20
Step 5: Since the base of the exponential expression is 5, which is greater
than 1, there are no real solutions for x. Thus, the equation 4(5x)−20 = 5x+1
has no solution in the real number system.
16
Question 25
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33back into the equation and solve for x:
32x−1= 33
Step 3: By the property of equality for exponential functions, we can equate
the exponents and solve for x:
2x−1 = 3
Step 4: Add 1 to both sides of the equation:
2x= 4
Step 5: Divide by 2 to solve for x:
x= 2
Step 6: Therefore, the solution to the exponential equation 32x−1= 27 is
x= 2.
Question 26
Question
Solve the exponential equation 52x−1= 125.
Solution
Step 1: Rewrite the equation using the same base on both sides.
52x−1= 125
52x−1= 53
Step 2: Set the exponents equal to each other.
2x−1 = 3
17
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 4: Verify the solution. Substitute x= 2 back into the original equation:
52(2)−1= 125
54−1= 125
53= 125
125 = 125
Therefore, the solution to the exponential equation is x= 2.
Question 27
Question
Solve for xin the equation 32x= 27.
Solution
Step 1: Rewrite 27 using the base 3. Step 2: Use the properties of exponents
to solve for x.
Question 28
Question
Solve the following exponential equation for x:
23x−1= 8x+2
Solution
Step 1: Rewrite 8as a power of 2.
23x−1= (23)x+2
Step 2: Simplify the right side of the equation.
23x−1= 23(x+2)
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
3x−1 = 3(x+ 2)
18
Step 4: Expand and solve for x.
3x−1 = 3x+ 6
−1 = 6
Step 5: Since we reached a false statement, there is no solution to the equa-
tion.
Question 29
Question
Solve the exponential equation for x:32x+1 −10·3x+9 = 0. Round your answer
to three decimal places if necessary.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation to solve for
3x. Let y= 3x. Then, we have y2−10y+ 9 = 0.
Step 2: Now, let’s solve the quadratic equation y2−10y+ 9 = 0 using the
quadratic formula: y=−(−10)±√(−10)2−4(1)(9)
2(1) .
Step 3: Simplifying gives y=10±√100−36
2.
Step 4: Therefore, we have two possible values for y:y1=10+√64
2and
y2=10−√64
2.
Step 5: Solving for y1gives y1=10+8
2= 9, and solving for y2gives y2=
10−8
2= 1.
Step 6: Recall that y= 3x. Therefore, we have two possible exponential
equations to solve: 3x= 9 and 3x= 1.
Step 7: Solving 3x= 9 gives x= log3(9) = 2, and solving 3x= 1 gives
x= log3(1) = 0.
Step 8: Thus, the solutions to the original equation 32x+1 −10 ·3x+ 9 = 0
are x= 2 and x= 0.
Question 30
Question
Solve the following exponential equation for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2:
23x−1= 23
19
Step 2: Since the bases are the same, we can equate the exponents:
3x−1 = 3
Step 3: Solve for x:
3x−1 = 3
3x= 4
x=4
3
Step 4: Therefore, the solution to the given exponential equation is x=4
3.
Question 31
Question
Solve the exponential equation: 32x+1 = 9.
Solution
Step 1: Rewrite 9 as a power of 3.
9 = 32
Step 2: Substitute 32back into the equation.
32x+1 = 32
Step 3: Set the exponents equal to each other.
2x+ 1 = 2
Step 4: Solve for x.
2x= 2 −1
2x= 1
x=1
2
Step 5: Check the solution by substituting x=1
2back into the original
equation.
32( 1
2)+1 = 32
32= 9
Since the left side equals the right side, the solution x=1
2is correct.
20
Question 32
Question
Suppose the population of a city can be modeled by the function P(t) = 8000 ·
e0.02t, where tis the number of years since 2020. Find the population of the
city in the year 2030.
Solution
Step 1: To find the population of the city in the year 2030, we need to substitute
t= 10 into the population function P(t).
P(10) = 8000 ·e0.02·10
Step 2: Calculate the exponential term.
P(10) = 8000 ·e0.2
Step 3: Evaluate e0.2using a calculator or by approximating it as follows:
e0.2≈1.221
Step 4: Substitute the value of e0.2back into the equation.
P(10) = 8000 ·1.221
Step 5: Multiply to find the population.
P(10) = 9768
Therefore, the population of the city in the year 2030 is 9768.
Question 33
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where
tis the number of years since the year 2010.
a) Determine the population of the city in the year 2025.
b) In what year will the population reach 10,000?
Solution
a) To determine the population in the year 2025, we need to find P(15) since
2025 is 15 years after 2010.
P(15) = 5000 ·e0.02·15
21
Step 1: Calculate the exponent inside the function.
0.02 ·15 = 0.3
Step 2: Substitute the exponent back into the equation and simplify.
P(15) = 5000 ·e0.3
Step 3: Calculate e0.3as an approximation.
e0.3≈1.3499
Step 4: Substitute the approximation back into the equation and find the
population.
P(15) = 5000 ·1.3499 = 6749.5
Therefore, the population of the city in the year 2025 is approximately
6749.5.
b) To find the year when the population reaches 10,000, we need to solve
the equation 5000 ·e0.02t= 10000.
Step 1: Divide both sides by 5000 to isolate the exponential term.
e0.02t= 2
Step 2: Take the natural logarithm of both sides to remove the base e.
ln(e0.02t)= ln(2)
Step 3: Use the property of logarithms to bring down the exponent.
0.02t·ln(e) = ln(2)
Step 4: Since ln(e) = 1, divide both sides by 0.02 to solve for t.
t=ln(2)
0.02 ≈34.66
Therefore, the population will reach 10,000 in the year 2044 (rounded up).
Question 34
Question
Solve the exponential equation 4x−1= 8.
22
Solution
Step 1: Rewrite 8as a power of 4using the property 43= 64.
Step 2: Substitute 8as 42in the equation.
(42)x−1= 43
Step 3: Simplify the left side using the property (am)n=am·n.
42(x−1) = 43
Step 4: Apply the power rule for exponents by multiplying the exponents.
42x−2= 43
Step 5: Since the bases are the same, set the exponents equal to each other.
2x−2 = 3
Step 6: Solve for xby isolating the variable.
2x= 5
x=5
2
Step 7: Therefore, the solution to the exponential equation 4x−1= 8 is
x=5
2.
Question 35
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where t
represents the number of years since the population was recorded. What is the
population after 10 years to the nearest whole number?
Solution
Step 1: Substitute t= 10 into the population function to find P(10). Step 2:
Calculate the population after 10 years by simplifying the expression. Step 3:
Round the final answer to the nearest whole number.
Step 1: Substitute t= 10 into the population function:
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression:
P(10) = 5000 ·e0.2
Step 3: Calculate the value of P(10):
P(10) ≈5000 ·1.221402758 ≈6107
Therefore, the population of the city after 10 years is approximately 6107
people.
23
Question 4
Question
Samantha invests $10,000 in a savings account with an annual interest rate
of 5%, compounded continuously. How long will it take for her investment to
double in value?
Solution
Step 1: First, we need to determine the continuous compound interest formula.
The formula for the amount of money after tyears in an account with principal
P, annual interest rate r, and compounded continuously is given by:
A=P·ert
where Ais the amount of money in the account after tyears, Pis the principal
amount, ris the annual interest rate, and eis the base of the natural logarithm
system.
Step 2: Since Samantha wants her investment to double, the amount of
money in her account after doubling will be 2P, where Pis the initial principal
amount. Therefore, we can write:
2P=P·e0.05t
Step 3: Now, we can solve for tby dividing both sides of the equation by P
and then taking the natural logarithm of both sides to solve for t:
ln(2) = 0.05t
Step 4: Next, we isolate tby dividing both sides by 0.05:
ln(2)
0.05 =t
Step 5: Now, we can calculate the value of t:
t≈ln(2)
0.05
Step 6: Using a calculator, we find:
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 in value when the interest is compounded continuously.
3
Question 5
Question
Let f(x)=3xand g(x) = log5(x). Determine the value of xthat satisfies the
equation f(g(x)) = 81.
Solution
Step 1: Start by substituting g(x)into f(x)and set the expression equal to 81.
f(g(x)) = f(log5(x))
= 3log5(x)
=x
Step 2: Set x= 81 and solve for x.
x= 81
log5(x) = log5(81)
log5(x) = log5(54)
log5(x) = 4
x= 54
x= 625
Therefore, the value of xthat satisfies the equation f(g(x)) = 81 is x= 625.
Question 6
Question
Suppose a certain investment grows according to the formula A(t) = 5000·e0.08t,
where A(t)represents the amount of money in the investment after tyears.
a) Calculate the amount of money in the investment after 5 years. Round
your answer to the nearest dollar.
b) Determine how many years it will take for the investment to double in
value.
Solution
a) To find the amount of money in the investment after 5 years, we evaluate
A(5) in the given formula.
A(5) = 5000 ·e0.08·5
Step 1: Calculate 0.08 ·5 = 0.4.
A(5) = 5000 ·e0.4
4
Step 2: Evaluate e0.4.
A(5) = 5000 ·e0.4
= 5000 ·1.491824693
≈7459.12
Therefore, the amount of money in the investment after 5 years is approxi-
mately $7459.
b) To determine how many years it will take for the investment to double in
value, we need to find the value of twhen A(t) = 2 ·A(0) = 2 ·5000 = 10000.
2·A(0) = 5000 ·e0.08t
10000 = 5000 ·e0.08t
2 = e0.08t
Step 1: Take the natural logarithm on both sides to solve for t.
ln(2) = ln(e0.08t)
ln(2) = 0.08t
t=ln(2)
0.08
Step 2: Calculate t.
t=ln(2)
0.08
≈0.693147181
0.08
≈8.664
Therefore, it will take approximately 8.664 years for the investment to double
in value.
Question 7
Question
Samantha invests $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much money will be in the account after 10 years?
Solution
Step 1: First, we need to determine the formula for the amount of money in the
account after tyears when interest is compounded continuously. This formula
is given by:
A=P·ert
5
where: - Ais the final amount in the account, - Pis the principal amount (initial
investment), - ris the interest rate, - tis the time in years, - eis Euler’s number
(≈2.71828).
Step 2: Substituting P= 5000,r= 0.035 (since 3.5% is equivalent to 0.035
as a decimal), and t= 10 into the formula, we get:
A= 5000 ·e0.035·10
Step 3: Calculate the exponent 0.035 ·10 = 0.35:
A= 5000 ·e0.35
Step 4: Using the approximation e0.35 ≈1.419067:
A≈5000 ·1.419067
Step 5: Calculate the final amount in the account after 10 years:
A≈7095.335
Therefore, there will be approximately $7095.34 in the account after 10 years
when interest is compounded continuously.
Question 8
Question
Let f(x) = 3xand g(x) = log2(x). Find the value of f(g(8)).
Solution
Step 1: First, we need to find g(8) by plugging x= 8 into g(x).
Step 1: g(8) = log2(8)
Step 2: Since 23= 8, we can rewrite g(8) as:
Step 2: g(8) = log2(8) = log2(23) = 3
Step 3: Now, we need to find f(g(8)) by plugging x= 3 into f(x).
Step 3: f(g(8)) = f(3) = 33
Step 4: Finally, calculate 33.
Step 4: 33= 27
Therefore, the value of f(g(8)) is 27 .
6
Question 9
Question
Samantha has invested $10,000 in a savings account that offers 4% interest
compounded continuously. How much will she have in the account after 5 years?
Solution
Step 1: Use the formula for continuously compounded interest:
A=P ert
where: - Ais the amount of money accumulated after tyears, - Pis the prin-
cipal amount (initial investment), - ris the annual interest rate (decimal), -
tis the time the money is invested for (in years), and - eis Euler’s number
(approximately equal to 2.71828).
Step 2: Plug in the given values. Given: - P= $10,000, - r= 0.04 (4% as a
decimal), - t= 5 years.
We have:
A= 10000 ×e0.04×5
Step 3: Calculate the exponent first:
0.04 ×5 = 0.2
Step 4: Plug this back into the formula and solve for A:
A= 10000 ×e0.2
Step 5: Calculate e0.2≈1.221402 (rounding to six decimal places).
Step 6: Multiply to find the final amount:
A≈10000 ×1.221402 ≈$12,214.02
Therefore, Samantha will have approximately $12,214.02 in the account after
5 years.
Question 10
Question
A certain radioactive substance decays according to the formula A(t) = A0e−kt,
where A(t)represents the amount of substance at time t,A0is the initial
amount, and kis a positive constant. If 50 grams of a substance decays to
40 grams after 2 days, find the value of kfor this substance.
7
Solution
Step 1: We are given that the initial amount of substance is A0= 50 grams,
the amount after 2 days is A(2) = 40 grams, and the time is t= 2 days. We
can plug these values into the radioactive decay formula to get an equation to
solve for k:
40 = 50e−2k
Step 2: Divide both sides of the equation by 50 to isolate the exponential
term:
0.8 = e−2k
Step 3: Take the natural logarithm of both sides to solve for k:
ln(0.8) = ln(e−2k)
Step 4: Use the property of logarithms ln(ex) = xto simplify the equation:
ln(0.8) = −2k
Step 5: Solve for kby dividing both sides by −2:
k=ln(0.8)
−2
Step 6: Use a calculator to approximate the value of k:
k≈ln(0.8)
−2≈−0.2231
−2≈0.1115
Therefore, the value of kfor this radioactive substance is approximately
0.1115.
Question 11
Question
Solve the exponential equation 32x−3 = 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute this into the original equation:
32x−3 = 33
Step 3: Use the properties of exponents to simplify the equation:
32x= 33+ 3
8
32x= 34
Step 4: Set the exponents equal to each other:
2x= 4
Step 5: Solve for x:
x=4
2
x= 2
Therefore, the solution to the equation 32x−3 = 27 is x= 2.
Question 12
Question
Solve the exponential equation 32x= 27.
Solution
Step 1: Express 27 as a power of 3.
27 = 33
Step 2: Set the exponent 2xequal to 3.
2x= 3
Step 3: Solve for xby dividing both sides by 2.
x=3
2
Therefore, the solution to the exponential equation 32x= 27 is x=3
2.
Question 13
Question
Let f(x) = 3xand g(x) = log3(x). Determine the domain of f(g(x)).
9
Solution
Step 1: To find the domain of f(g(x)), we need to consider the restrictions on
the composition function.
Step 2: Since g(x) = log3(x), we know that g(x)is defined only for positive
values of x, i.e., x > 0.
Step 3: Now, we need to substitute g(x)into f(x)to find f(g(x)).
f(g(x)) = f(log3(x)) = 3log3(x)
Step 4: Remember that 3log3(x)=x.
Step 5: Therefore, the domain of f(g(x)) is all positive real numbers. In
interval notation, the domain is (0,∞).
Question 14
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where t
represents the number of years since the population started being tracked. Find
the population of the city after 10 years.
Solution
Step 1: Substitute t= 10 into the formula P(t) = 5000 ·1.02tto find the
population after 10 years.
P(10) = 5000 ·1.0210
= 5000 ·1.218994
= 6094.97
Step 2: Round the population to the nearest whole number since a popula-
tion count must be a whole number. Therefore, the population of the city after
10 years is approximately 6095.
Question 15
Question
Solve the exponential equation 2x−3·2x−1−4 = 0.
Solution
Step 1: Let’s rewrite the equation by factoring out 2x−1:
2x−3·2x−1−4 = 0
10
2x−1(2 −3) −4 = 0
2x−1−4 = 0
Step 2: Now, we have a simple exponential equation. Let’s solve for 2x−1:
2x−1= 4
Step 3: Rewrite 4as 22:
2x−1= 22
Step 4: Since the bases are the same, we can equate the exponents:
x−1 = 2
Step 5: Solve for x:
x= 2 + 1
x= 3
Step 6: Therefore, the solution to the given exponential equation is x= 3.
Question 16
Question
Suppose an investment of $5000 is made in an account that pays 3% interest,
compounded continuously. How long will it take for the investment to double
in value?
Solution
Let’s denote the initial investment amount as P= $5000, the annual interest
rate as r= 0.03 (3%), and the time required to double the investment as t.
We use the continuous compounding formula for exponential growth:
A=P ert
where Ais the final amount, Pis the principal amount, eis the base of the
natural logarithm, ris the interest rate, and tis time in years.
Step 1: We want the investment to double, so the final amount will be
2P= 2 ×5000 = $10000. Therefore, our equation becomes:
10000 = 5000e0.03t
Step 2: Divide both sides of the equation by 5000 to isolate the exponential
term:
2 = e0.03t
11
Step 3: Take the natural logarithm (ln) of both sides to solve for t:
ln(2) = ln(e0.03t)
ln(2) = 0.03t
t=ln(2)
0.03
Step 4: Now, calculate the value of t:
t=ln(2)
0.03 ≈0.6931
0.03 ≈23.1045
Therefore, it will take approximately 23.1 years for the investment to double
in value when continuously compounded at 3% interest.
Question 17
Question
Solve the exponential equation 3x−4= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33for 27 in the equation:
3x−4= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other:
x−4 = 3
Step 4: Solve for x:
x= 3 + 4
x= 7
Step 5: Therefore, the solution to the exponential equation 3x−4= 27 is
x= 7.
Question 18
Question
Solve for x:e2x−5ex+ 6 = 0.
12
Solution
Step 1: Let u=ex. Then, the equation becomes u2−5u+ 6 = 0.
Step 2: Factor the quadratic equation: (u−2)(u−3) = 0.
Step 3: Set each factor to zero: u−2 = 0 or u−3 = 0.
Step 4: Solve for uin each case: For u−2 = 0:u= 2. For u−3 = 0:u= 3.
Step 5: Recall that u=ex.
Step 6: Solve for x: For u= 2:ex= 2 =⇒x= ln(2). For u= 3:
ex= 3 =⇒x= ln(3).
Therefore, the solutions to the equation e2x−5ex+ 6 = 0 are x= ln(2) and
x= ln(3).
Question 19
Question
Suppose an investment of $5000 grows exponentially with a continuous interest
rate of 6% per year. How much will the investment be worth after 10 years?
Solution
Step 1: Recall the formula for compound interest in continuous compounding:
A=P·ert
where: A= the amount of money accumulated after tyears, including interest
P= the principal amount (initial investment) r= the annual interest rate
(expressed as a decimal) t= the time the money is invested for
Step 2: Substitute the given values into the formula: The principal amount,
P= $5000 The annual interest rate, r= 0.06 The time period, t= 10 years
Step 3: Plug the values into the formula and calculate:
A= 5000 ·e0.06·10
A= 5000 ·e0.6
A≈5000 ·1.82212
A≈9110.6
Therefore, the investment will be worth approximately $9110.6after 10
years.
Question 20
Question
The population of a city can be modeled by the function P(t) = 2500 ·1.03t,
where trepresents the number of years since the year 2020. At what rate is the
population growing in 2025?
13
Solution
Step 1: Calculate the population in 2025 using the given function.
Step 2: Find the population in 2026 by plugging in t= 6 in the function and
calculate the difference between the populations for 2025 and 2026.
Step 3: Calculate the rate of population growth by taking the difference between
the populations and dividing it by 1 year.
Step 1: Calculate the population in 2025:
P(5) = 2500 ·1.035= 2500 ·1.159274 = 2898.185.
The population in 2025 is approximately 2898.185.
Step 2: Calculate the population in 2026:
P(6) = 2500 ·1.036= 2500 ·1.19101622 = 2977.54.
The population in 2026 is approximately 2977.54.
The rate of population growth between 2025 and 2026 is:
2977.54 −2898.185 = 79.355.
Step 3: Calculate the rate of population growth:
79.355
1= 79.355.
Therefore, the rate at which the population is growing in 2025 is approxi-
mately 79.355 people per year.
Question 21
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: Set the equation equal to the new representation of 27.
32x+1 = 33
Step 3: Since the bases are the same, set the exponents equal to each other.
2x+ 1 = 3
14
Step 4: Solve for xby isolating the variable.
2x+ 1 = 3
2x= 2
x= 1
Step 5: Check the solution by substituting x= 1 back into the original
equation.
32(1)+1 = 33
33= 27
Therefore, the solution to the equation is x= 1.
Question 22
Question
Solve the exponential equation 3x−1+ 3x−2= 10.
Solution
Step 1: Let’s rewrite the equation in terms of a single base:
3x−1+ 3x−2= 3 ·3x−2+ 3x−2= 3 ·3x−2+ 3x−2= 3x−2(3 + 1) = 4 ·3x−2
Step 2: Rewrite the equation as 4·3x−2= 10.
Step 3: Divide by 4 to isolate 3x−2:
3x−2=10
4=5
2
Step 4: Write the right-hand side as 31−log32:
Step 5: Since both sides have the same base, we can set the exponents equal
to each other:
x−2 = 1 −log32
Step 6: Add 2 to both sides to solve for x:
x= 1 + 2 −log32 = 3 −log32
Hence, the solution to the exponential equation 3x−1+ 3x−2= 10 is x=
3−log32.
Question 23
Question
Samantha invested $10,000 in an account that earns 3% compounded continu-
ously. How much will the investment be worth after 5 years?
15
Solution
Step 1: Identify the formula for continuous compounding which is given by:
A=P·ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= the annual interest rate (in decimal form), and
t= the time the money is invested for.
Step 2: Plug in the given values into the formula: P= $10,000,r= 0.03
(3% as a decimal), and t= 5 years.
Step 3: Calculate the amount accumulated after 5 years using the formula.
A= $10,000 ·e0.03·5
Step 4: Simplify the expression and calculate the final amount.
A= $10,000 ·e0.15
A≈$10,000 ·1.161834
A≈$11,618.34
Therefore, the investment will be worth approximately $11,618.34 after 5
years when compounded continuously.
Question 24
Question
Solve the exponential equation for x:4(5x)−20 = 5x+1.
Solution
Step 1: Let’s rewrite the given equation by recognizing that 5x+1 = 5 ·5x.
4·5x−20 = 5 ·5x
Step 2: Subtract 5xfrom both sides of the equation to isolate 5xon one side.
4·5x−5·5x= 20
Step 3: Combine like terms on the left side of the equation.
4·5x−5·5x= 20
−1·5x= 20
Step 4: Divide both sides by −1to solve for 5x.
5x=−20
Step 5: Since the base of the exponential expression is 5, which is greater
than 1, there are no real solutions for x. Thus, the equation 4(5x)−20 = 5x+1
has no solution in the real number system.
16
Question 25
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
Step 2: Substitute 33back into the equation and solve for x:
32x−1= 33
Step 3: By the property of equality for exponential functions, we can equate
the exponents and solve for x:
2x−1 = 3
Step 4: Add 1 to both sides of the equation:
2x= 4
Step 5: Divide by 2 to solve for x:
x= 2
Step 6: Therefore, the solution to the exponential equation 32x−1= 27 is
x= 2.
Question 26
Question
Solve the exponential equation 52x−1= 125.
Solution
Step 1: Rewrite the equation using the same base on both sides.
52x−1= 125
52x−1= 53
Step 2: Set the exponents equal to each other.
2x−1 = 3
17
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 4: Verify the solution. Substitute x= 2 back into the original equation:
52(2)−1= 125
54−1= 125
53= 125
125 = 125
Therefore, the solution to the exponential equation is x= 2.
Question 27
Question
Solve for xin the equation 32x= 27.
Solution
Step 1: Rewrite 27 using the base 3. Step 2: Use the properties of exponents
to solve for x.
Question 28
Question
Solve the following exponential equation for x:
23x−1= 8x+2
Solution
Step 1: Rewrite 8as a power of 2.
23x−1= (23)x+2
Step 2: Simplify the right side of the equation.
23x−1= 23(x+2)
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
3x−1 = 3(x+ 2)
18
Step 4: Expand and solve for x.
3x−1 = 3x+ 6
−1 = 6
Step 5: Since we reached a false statement, there is no solution to the equa-
tion.
Question 29
Question
Solve the exponential equation for x:32x+1 −10·3x+9 = 0. Round your answer
to three decimal places if necessary.
Solution
Step 1: Let’s rewrite the equation in terms of a quadratic equation to solve for
3x. Let y= 3x. Then, we have y2−10y+ 9 = 0.
Step 2: Now, let’s solve the quadratic equation y2−10y+ 9 = 0 using the
quadratic formula: y=−(−10)±√(−10)2−4(1)(9)
2(1) .
Step 3: Simplifying gives y=10±√100−36
2.
Step 4: Therefore, we have two possible values for y:y1=10+√64
2and
y2=10−√64
2.
Step 5: Solving for y1gives y1=10+8
2= 9, and solving for y2gives y2=
10−8
2= 1.
Step 6: Recall that y= 3x. Therefore, we have two possible exponential
equations to solve: 3x= 9 and 3x= 1.
Step 7: Solving 3x= 9 gives x= log3(9) = 2, and solving 3x= 1 gives
x= log3(1) = 0.
Step 8: Thus, the solutions to the original equation 32x+1 −10 ·3x+ 9 = 0
are x= 2 and x= 0.
Question 30
Question
Solve the following exponential equation for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2:
23x−1= 23
19
Step 2: Since the bases are the same, we can equate the exponents:
3x−1 = 3
Step 3: Solve for x:
3x−1 = 3
3x= 4
x=4
3
Step 4: Therefore, the solution to the given exponential equation is x=4
3.
Question 31
Question
Solve the exponential equation: 32x+1 = 9.
Solution
Step 1: Rewrite 9 as a power of 3.
9 = 32
Step 2: Substitute 32back into the equation.
32x+1 = 32
Step 3: Set the exponents equal to each other.
2x+ 1 = 2
Step 4: Solve for x.
2x= 2 −1
2x= 1
x=1
2
Step 5: Check the solution by substituting x=1
2back into the original
equation.
32( 1
2)+1 = 32
32= 9
Since the left side equals the right side, the solution x=1
2is correct.
20
Question 32
Question
Suppose the population of a city can be modeled by the function P(t) = 8000 ·
e0.02t, where tis the number of years since 2020. Find the population of the
city in the year 2030.
Solution
Step 1: To find the population of the city in the year 2030, we need to substitute
t= 10 into the population function P(t).
P(10) = 8000 ·e0.02·10
Step 2: Calculate the exponential term.
P(10) = 8000 ·e0.2
Step 3: Evaluate e0.2using a calculator or by approximating it as follows:
e0.2≈1.221
Step 4: Substitute the value of e0.2back into the equation.
P(10) = 8000 ·1.221
Step 5: Multiply to find the population.
P(10) = 9768
Therefore, the population of the city in the year 2030 is 9768.
Question 33
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where
tis the number of years since the year 2010.
a) Determine the population of the city in the year 2025.
b) In what year will the population reach 10,000?
Solution
a) To determine the population in the year 2025, we need to find P(15) since
2025 is 15 years after 2010.
P(15) = 5000 ·e0.02·15
21
Step 1: Calculate the exponent inside the function.
0.02 ·15 = 0.3
Step 2: Substitute the exponent back into the equation and simplify.
P(15) = 5000 ·e0.3
Step 3: Calculate e0.3as an approximation.
e0.3≈1.3499
Step 4: Substitute the approximation back into the equation and find the
population.
P(15) = 5000 ·1.3499 = 6749.5
Therefore, the population of the city in the year 2025 is approximately
6749.5.
b) To find the year when the population reaches 10,000, we need to solve
the equation 5000 ·e0.02t= 10000.
Step 1: Divide both sides by 5000 to isolate the exponential term.
e0.02t= 2
Step 2: Take the natural logarithm of both sides to remove the base e.
ln(e0.02t)= ln(2)
Step 3: Use the property of logarithms to bring down the exponent.
0.02t·ln(e) = ln(2)
Step 4: Since ln(e) = 1, divide both sides by 0.02 to solve for t.
t=ln(2)
0.02 ≈34.66
Therefore, the population will reach 10,000 in the year 2044 (rounded up).
Question 34
Question
Solve the exponential equation 4x−1= 8.
22
Solution
Step 1: Rewrite 8as a power of 4using the property 43= 64.
Step 2: Substitute 8as 42in the equation.
(42)x−1= 43
Step 3: Simplify the left side using the property (am)n=am·n.
42(x−1) = 43
Step 4: Apply the power rule for exponents by multiplying the exponents.
42x−2= 43
Step 5: Since the bases are the same, set the exponents equal to each other.
2x−2 = 3
Step 6: Solve for xby isolating the variable.
2x= 5
x=5
2
Step 7: Therefore, the solution to the exponential equation 4x−1= 8 is
x=5
2.
Question 35
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where t
represents the number of years since the population was recorded. What is the
population after 10 years to the nearest whole number?
Solution
Step 1: Substitute t= 10 into the population function to find P(10). Step 2:
Calculate the population after 10 years by simplifying the expression. Step 3:
Round the final answer to the nearest whole number.
Step 1: Substitute t= 10 into the population function:
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression:
P(10) = 5000 ·e0.2
Step 3: Calculate the value of P(10):
P(10) ≈5000 ·1.221402758 ≈6107
Therefore, the population of the city after 10 years is approximately 6107
people.
23
