MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 6
Liberty University
Question 1
Question
Samantha invests $10,000 in an account that pays 6% interest, compounded
annually. How much money will be in the account after 5 years? Round your
answer to the nearest dollar.
Solution
Step 1: The formula for compound interest is given by the formula
A=P(1 + r
n)nt
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). - nis the number of times that interest is
compounded per year. - tis the time the money is invested for in years.
In this case, P= $10,000,r= 0.06,n= 1, and t= 5.
Step 2: Substitute the given values into the compound interest formula and
calculate the amount of money accumulated after 5 years:
A= 10000 (1 + 0.06
1)1×5
A= 10000 ×(1.06)5
Step 3: Calculate the value of (1.06)5:
(1.06)5≈1.338225
Step 4: Multiply this value by 10,000 to find the amount of money in the
account after 5 years:
A≈10000 ×1.338225
A≈13382.25
Therefore, after 5 years, there will be approximately $13,382 in the account.
Question 2
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since the initial population count.
Determine the population of the city after 10 years, rounded to the nearest
whole number.
Solution
Step 1: Substitute t= 10 into the population function P(t)to find the popula-
tion after 10 years.
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210.
1.0210 = 1.218391]
Step 3: Substitute this value back into the population equation.
P(10) = 5000 ·1.218391
Step 4: Multiply to find the population after 10 years.
P(10) ≈6091
Therefore, the population of the city after 10 years is approximately 6091.
Question 3
Question
Samantha invests $5000 in a savings account that earns 3% interest compounded
quarterly. How much will the investment be worth after 10 years?
2
Solution
Let’s start by setting up the formula for compound interest, which is given by:
A=P(1 + r
n)nt
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). - nis the number of times that interest is
compounded per year. - tis the time the money is invested for in years.
In this case: - P= 5000 (the initial investment). - r= 0.03 (3% annual
interest rate as a decimal). - n= 4 (quarterly compounding). - t= 10 years.
Substitute these values into the compound interest formula to find out the
final amount after 10 years.
Step 1: Calculate the amount using the compound interest formula.
A= 5000 (1 + 0.03
4)4×10
Step 2: Simplify the expression inside the parentheses.
A= 5000 (1 + 0.0075)40
Step 3: Add the numbers inside the parentheses.
A= 5000 ×1.007540
Step 4: Calculate the value of 1.007540.
A≈5000 ×1.3498
Step 5: Multiply to find the final amount.
A≈6749.02
Therefore, the investment will be worth approximately $6749.02 after 10
years.
Question 4
Question
Suppose a population of bacteria doubles every 4 hours. If there are initially
100 bacteria, determine the exponential function that models the population of
bacteria after thours.
3
Solution
Step 1: Let P(t)be the population of bacteria after thours. Since the population
doubles every 4 hours, we can express P(t)as P(t) = 100 ·2(t/4).
Thus, the exponential function that models the population of bacteria after
thours is P(t) = 100 ·2(t/4).
Question 5
Question
A bacteria culture starts with 1000 bacteria and doubles in size every 3 hours.
Write an exponential growth model to represent the number of bacteria after t
hours, and determine how many bacteria there will be after 9 hours.
Solution
Step 1: To write the exponential growth model, we start with the general ex-
ponential function P(t) = P0·rt, where: - P(t)is the population after thours,
-P0is the initial population, - ris the growth rate per unit of time, and - tis
the time in hours.
Given that the bacteria doubles in size every 3 hours, the growth rate ris 2
(since doubling corresponds to multiplying by 2). The initial population P0is
1000. Therefore, the exponential growth model for this scenario is:
P(t) = 1000 ·2t/3
Step 2: To determine the number of bacteria after 9 hours, we substitute
t= 9 into the exponential growth model:
P(9) = 1000 ·29/3
P(9) = 1000 ·23
P(9) = 1000 ·8
P(9) = 8000
After 9 hours, there will be 8000 bacteria in the culture.
Question 6
Question
Suppose a population of insects is growing exponentially, with an initial popu-
lation of 100 insects and a growth rate of 20
4
Solution
Step 1: The exponential growth model for this scenario can be expressed as:
P(t) = P0·(1 + r)t
where: - P(t)is the population after tdays, - P0is the initial population, - ris
the growth rate per day (as a decimal), and - tis the number of days.
Step 2: Substituting the given values into the formula, we have:
P(t) = 100 ·(1 + 0.20)t
Step 3: Simplifying the expression, we get:
P(t) = 100 ·(1.20)t
P(t) = 100 ·1.20t
Therefore, the exponential growth model for this scenario is P(t) = 100 ·
1.20t.
Question 7
Question
Solve the following exponential equation: 3x+1 −3x= 12.
Solution
Step 1: We start by simplifying the left side of the equation using the properties
of exponents.
Step 2: Since 3x+1 = 3 ·3x, we rewrite the equation as 3·3x−3x= 12.
Step 3: Combining like terms, we get 2·3x= 12.
Step 4: Divide by 2 to isolate 3x:3x= 6.
Step 5: Now, we can rewrite 6as 31·2.
Step 6: By the properties of exponents, we can rewrite the equation as
3x= 31·3log32.
Step 7: Equating the exponents, we have x= 1 + log32.
Therefore, the solution to the equation 3x+1 −3x= 12 is x= 1 + log32.
Question 8
Question
Samantha invests $5000 in a savings account that pays an annual interest rate
of 4.5%, compounded continuously. How long will it take for her investment to
double in value?
5
Solution
Let A(t)be the amount of money in the account after tyears. We know that
the formula for continuously compounded interest is given by A(t) = P·ert,
where Pis the principal amount, ris the annual interest rate, and tis the time
in years.
Step 1: Substitute the given values into the formula. The principal amount
Pis $5000, the annual interest rate ris 0.045 (4.5% as a decimal), and we want
to find the time tit takes for the investment to double, so A(t) = 2P. Therefore,
the equation becomes:
2·5000 = 5000 ·e0.045t
Step 2: Solve for t. Dividing both sides by 5000, we get:
2 = e0.045t
Taking the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
Using the property ln(ex) = x:
ln(2) = 0.045t
Now, solving for t:
t=ln(2)
0.045 ≈0.6931
0.045 ≈15.47
Therefore, it will take about 15.47 years for Samantha’s investment to double
in value when compounded continuously.
Question 9
Question
Solve the exponential equation for x:2x−3−2x−1= 4.
Solution
To solve the given exponential equation, we need to manipulate the equation in
a way that allows us to isolate the variable x.
Step 1: Rewrite the equation using the properties of exponents.
2x−3−2x−1= 4.
Step 2: Recognize that 2x−3=1
2·2xand 2x−1= 2 ·2x.
1
2·2x−2·2x= 4.
6
Step 3: Combine like terms on the left side.
(1
2−2)·2x= 4.
Step 4: Simplify the left side of the equation.
−3
2·2x= 4.
Step 5: Solve for xby isolating 2x.
2x=−4
3
2
=−8
3.
Step 6: Rewrite the equation in exponential form.
x= log2(−8
3).
Since the logarithm of a negative number is undefined for real numbers, there
is no solution to the original exponential equation.
Question 10
Question
Suppose that an investment of $5000 grows according to the formula A= 5000 ·
e0.04t, where Ais the amount of money after tyears. How long will it take for
the investment to double?
Solution
Step 1: To find the time it takes for the investment to double, we need to solve
the equation 2·5000 = 5000 ·e0.04t.
Step 2: Simplifying the equation, we have 10000 = 5000 ·e0.04t.
Step 3: Dividing both sides by 5000, we get 2 = e0.04t.
Step 4: Taking the natural logarithm of both sides, we have ln(2) = ln(e0.04t).
Step 5: Using the property ln(ex) = x, we can simplify further to ln(2) =
0.04t.
Step 6: Next, we solve for tby dividing by 0.04:t=ln(2)
0.04 .
Step 7: Calculating this expression, t≈0.6931
0.04 ≈17.33 years.
Step 8: Therefore, it will take approximately 17.33 years for the investment
to double.
Question 11
Question
Solve the exponential equation 23x−1= 8.
7
Solution
Step 1: Rewrite 8as a power of 2.
23= 8
Step 2: Substitute 8as 23in the equation.
23x−1= 23
Step 3: Set the exponents equal to each other.
3x−1 = 3
Step 4: Solve for x.
3x= 4
x=4
3
Step 5: Check the solution by substituting x=4
3back into the original
equation.
23( 4
3)−1= 8
24−1= 8
23= 8
8 = 8
Therefore, the solution to the exponential equation is x=4
3.
Question 12
Question
Samantha invests $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much will Samantha have in the account after 10
years?
Solution
Let’s denote the amount of money in the account after tyears as A(t). We can
use the formula for continuous compounding:
A(t) = P·ert
where: - Pis the principal amount (initial investment), - ris the annual
interest rate (in decimal form), - tis the time in years, and - eis Euler’s number,
approximately equal to 2.71828.
8
In this case, P= $5000,r= 0.035, and t= 10. Plugging these values into
the formula, we get:
Step 1:
A(t) = 5000 ·e0.035·10
Step 2:
A(10) = 5000 ·e0.35
Step 3:
A(10) ≈5000 ·1.419067
Step 4:
A(10) ≈7095.335
Therefore, Samantha will have approximately $7095.34 in the account after
10 years.
Question 13
Question
Solve for xin the equation log2(2x+ 5) + log2(x−1) = 3.
Solution
Step 1: Rewrite the equation using the properties of logarithms.
log2(2x+ 5) + log2(x−1) = 3
log2((2x+ 5)(x−1)) = 3
Step 2: Remove the logarithm by expressing the equation in exponential
form.
23= (2x+ 5)(x−1)
8 = 2x2+ 3x−5
Step 3: Rearrange the equation into standard form by setting it equal to
zero.
2x2+ 3x−13 = 0
Step 4: Solve the quadratic equation using the quadratic formula x=
−b±√b2−4ac
2a. Using the values a= 2,b= 3, and c=−13:
x=−3±√32−4(2)(−13)
2(2)
x=−3±√9 + 104
4
x=−3±√113
4
Therefore, the solutions for xare x=−3+√113
4and x=−3−√113
4.
9
Question 14
Question
Samantha invests $10,000 in an account that pays 4.5% interest compounded
continuously. How long will it take for her investment to double?
Solution
Step 1: To find the time it takes for Samantha’s investment to double, we need
to use the formula for continuously compounded interest:
A=P·ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= the interest rate (decimal form), and t= the
time the money is invested for.
Step 2: Since Samantha wants to double her investment, we can set up the
equation as follows:
2·P=P·ert
Step 3: Substitute the given values into the equation:
2·10,000 = 10,000 ·e0.045t
Step 4: Simplify the equation:
20,000 = 10,000 ·e0.045t
Step 5: Divide both sides by 10,000 to solve for e0.045t:
2 = e0.045t
Step 6: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
Step 7: Remember that ln(ex) = x, so we have:
ln(2) = 0.045t
Step 8: Solve for tby dividing by 0.045:
t=ln(2)
0.045
Step 9: Using a calculator to approximate the value:
t≈0.6931
0.045 ≈15.47 years
Answer: It will take approximately 15.47 years for her investment to double
when earning 4.5% interest compounded continuously.
10
Question 15
Question
Suppose an investment account starts with an initial deposit of $5000 and earns
3% interest compounded annually. How long will it take for the account balance
to reach $8000?
Solution
Step 1: First, we need to determine the exponential growth model for this
scenario. The formula for compound interest is given by:
A=P(1 + r)t
where: - Ais the final amount in the account - Pis the principal amount (initial
deposit) - ris the annual interest rate in decimal form - tis the time the money
is invested for in years
Step 2: Substituting the given values into the formula:
8000 = 5000(1 + 0.03)t
Step 3: Divide both sides by 5000 to isolate the exponential term:
8000
5000 = (1 + 0.03)t
1.6 = (1.03)t
Step 4: To solve for t, we will take the natural logarithm of both sides to
eliminate the exponent:
ln(1.6) = ln(1.03)t
Step 5: By the properties of logarithms, we can bring down the exponent t:
ln(1.6) = t·ln(1.03)
Step 6: Finally, we solve for tby dividing both sides by ln(1.03):
t=ln(1.6)
ln(1.03) ≈0.4700
0.0296 ≈15.88
Therefore, it will take approximately 15.88 years for the account balance to
reach $8000.
Question 16
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since 2020. Determine the approximate year
when the population of the city will reach 10,000 people.
11
Solution
Step 1: Let’s set up the equation with the given information:
10,000 = 5000 ·1.02t
Step 2: Divide both sides by 5000 to isolate the exponential term:
2 = 1.02t
Step 3: To solve for t, we need to express the equation in logarithmic form:
log1.02 2 = t
Step 4: Use a calculator to compute log1.02 2:
t≈34.3
Step 5: This means that it will take approximately 34.3 years since 2020 for
the population to reach 10,000 people.
Step 6: Finally, to determine the year when this population will be reached,
add 34.3 to 2020:
2020 + 34.3≈2054.3
Therefore, the approximate year when the population of the city will reach
10,000 people is 2054.
Question 17
Question
Solve the following exponential equation for x:
4x+1 = 32
Solution
Step 1: Rewrite 32 as a power of 4. Step 2: Solve for xusing the properties of
exponents.
Step 1: We can rewrite 32 as a power of 4 by noticing that 32 = 42·2.
Therefore, 32 = 42·41= 42+1.
Step 2: Substitute 32 with 42+1 in the equation 4x+1 = 32:
4x+1 = 42+1
By the property of equality for exponents, we can equate the exponents:
x+ 1 = 2 + 1
Solving for xgives:
x= 2
Therefore, the solution to the equation 4x+1 = 32 is x= 2.
12
Question 18
Question
Solve the equation for x:32x= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 33= 27,we can rewrite 27 as 33.
Step 2: Rewrite the equation 32x= 27 with 27 expressed as 33.
32x= 33.
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x= 3.
Step 4: Solve for x.
2x= 3
x=3
2
x= 1.5.
Therefore, the solution to the equation 32x= 27 is x= 1.5.
Question 19
Question
Samantha invested $5000 in an account that earns 3% annual interest com-
pounded continuously. How long will it take for the investment to double in
value?
Solution
Step 1: The continuous compound interest formula is given by A=P ert, where:
-Ais the amount of money after tyears, - Pis the principal investment amount,
-ris the annual interest rate (as a decimal), and - tis the time in years.
Step 2: Since we want to know when the investment will double, we can set
up the equation 2P=P e0.03t, where P= 5000 and r= 0.03.
Step 3: Simplifying the equation, we get 2 = e0.03t.
Step 4: To solve for t, we take the natural logarithm of both sides: ln(2) =
ln(e0.03t).
Step 5: Using the properties of logarithms, we have ln(2) = 0.03t.
13
Step 6: Divide by 0.03 to solve for t:t=ln(2)
0.03 .
Step 7: Calculating the value of t, we have t≈ln(2)
0.03 ≈0.693
0.03 ≈23.1.
Step 8: Therefore, it will take approximately 23.1years for Samantha’s
investment to double in value.
Question 20
Question
Solve for xin the equation 23x−5 = 13.
Solution
Step 1: Add 5 to both sides of the equation to isolate the exponential term:
23x= 13 + 5 = 18
Step 2: Rewrite 18 as a power of 2:
23x= 2log218
Step 3: Set the exponents equal to each other:
3x= log218
Step 4: Solve for xby dividing both sides by 3:
x=log218
3
Step 5: Simplify the expression for xusing properties of logarithms:
x=log2(2log218)
3
x=log218
3
Therefore, the solution to the equation 23x−5 = 13 is x=log218
3.
Question 21
Question
Solve the following exponential equation for x:32x−3x+1 + 2 = 0.
14
Solution
Step 1: Let’s rewrite the equation in terms of 3x:
32x−3x+1 + 2 = 0
(3x)2−3·3x+ 2 = 0
Step 2: Now, let’s substitute y= 3xto simplify the equation:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
Step 4: Set each factor to zero and solve for y:
y−1 = 0 =⇒y= 1
y−2 = 0 =⇒y= 2
Step 5: Recall that y= 3x: For y= 1:
3x= 1
x= 0
For y= 2:
3x= 2
x= log32
Step 6: Therefore, the solutions to the equation are x= 0 and x= log32.
Question 22
Question
A certain population of bacteria doubles every 5 hours. If the initial population
is 1000 bacteria, find an exponential model expressing the population Pas a
function of time tin hours. Additionally, determine the population after 15
hours.
Solution
Step 1: To find the exponential model, we use the formula for exponential
growth: P(t) = P0·2(t/k), where P0is the initial population, kis the time it
takes for the population to double, and tis the time elapsed.
Step 2: Given that the population doubles every 5 hours, we have k= 5.
Also, the initial population P0= 1000. Therefore, the exponential model is
P(t) = 1000 ·2(t/5).
Step 3: To find the population after 15 hours, we substitute t= 15 into the
model: P(15) = 1000 ·2(15/5).
Step 4: Simplifying the expression gives P(15) = 1000 ·23= 1000 ·8 = 8000.
Therefore, the population after 15 hours is 8000 bacteria.
15
Question 23
Question
Suppose a certain investment grows according to the formula A(t) = 5000·e0.04t,
where A(t)represents the amount of money in the investment after tyears. Find
the rate of change of the investment after 4 years.
Solution
Step 1: Calculate the derivative of A(t)with respect to t.
A′(t) = d
dt (5000 ·e0.04t)
Step 2: Apply the chain rule to find the derivative of A(t).
A′(t) = 5000 ·d
dt (e0.04t)
Step 3: Differentiate e0.04twith respect to t.
A′(t) = 5000 ·0.04 ·e0.04t
A′(t) = 200e0.04t
Step 4: Evaluate the rate of change after 4 years by plugging in t= 4.
A′(4) = 200e0.04·4
A′(4) = 200e0.16
A′(4) ≈226.90
Therefore, the rate of change of the investment after 4 years is approximately
226.90.
Question 24
Question
Samantha deposits $5000 into a savings account that earns an annual interest
rate of 3%. How long will it take for the account balance to double if the interest
is compounded continuously?
16
Solution
Step 1: Let A(t)represent the amount of money in the savings account after t
years. We can use the formula for continuously compounded interest:
A(t) = P·ert,
where Pis the principal amount ($5000 in this case), ris the annual interest
rate (expressed as a decimal, so 0.03 for 3%), and tis the time in years.
Step 2: Since we want to find when the account balance doubles, we are
looking for tsuch that A(t) = 2 ·P= 2 ·$5000 = $10000. Substituting this into
the formula, we get:
10000 = 5000 ·e0.03t.
Step 3: Divide both sides by 5000 to isolate the exponential term:
10000
5000 =e0.03t.
Step 4: Simplify the left side to get:
2 = e0.03t.
Step 5: To solve for t, take the natural logarithm of both sides:
ln(2) = ln(e0.03t).
Step 6: Using the property ln(ex) = x, we get:
ln(2) = 0.03t.
Step 7: Finally, solve for tby dividing by 0.03:
t=ln(2)
0.03 ≈23.1.
Therefore, it will take approximately 23.1 years for the account balance to
double with continuously compounded interest.
Question 25
Question
Let f(x) = 3xand g(x) = log3(x). Determine the domain of the composite
function g(f(x)).
17
Solution
Step 1: Find the composite function g(f(x)):
g(f(x)) = log3(3x)
Step 2: Use the property of logarithmic functions loga(ax) = x:
g(f(x)) = x
Step 3: Determine the domain of g(f(x)). Since xis a real number, the
domain of g(f(x)) is also all real numbers. Therefore, the domain of g(f(x)) is
(−∞,∞).
Question 26
Question
Solve the exponential equation 32x−1= 81 for x.
Solution
Step 1: Rewrite 81 as a power of 3:81 = 34.
Step 2: Substitute 81 with 34in the equation: 32x−1= 34.
Step 3: Since the bases on both sides are the same, we can set the exponents
equal to each other: 2x−1 = 4.
Step 4: Solve the resulting linear equation for x:
2x−1 = 4
2x= 5
x=5
2
Step 5: Therefore, the solution to the exponential equation 32x−1= 81 is
x=5
2.
Question 27
Question
Samantha invests $10,000 in an account that pays 3.5% interest compounded
continuously. How long will it take for her investment to double in value?
18
Solution
Step 1: First, we need to find the formula for continuously compounded interest.
The formula is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the princi-
pal amount (initial investment), - ris the annual interest rate (expressed as a
decimal), - tis the time the money is invested for, - eis the base of the natural
logarithm, approximately equal to 2.71828.
Step 2: Since Samantha wants her investment to double, we set A= 2Pin
the formula:
2P=P·e0.035t
Step 3: Divide both sides by Pto simplify the equation:
2 = e0.035t
Step 4: Take the natural logarithm of both sides to solve for t:
ln 2 = ln e0.035t
ln 2 = 0.035t·ln e
ln 2 = 0.035t
Step 5: Solve for tby dividing both sides by 0.035:
t=ln 2
0.035
t≈0.6931
0.035
t≈19.802 years
Therefore, it will take approximately 19.802 years for Samantha’s investment
to double in value when compounded continuously at a rate of 3.5%.
Question 28
Question
A population of bacteria doubles every hour. If there are initially 100 bacteria,
how many bacteria will there be after 6 hours?
19
Solution
Step 1: Let’s denote the initial number of bacteria as P0= 100 and the growth
factor per hour as r= 2, since the population doubles every hour. The popula-
tion after thours can be represented by the exponential function P(t) = P0·rt.
Step 2: Substitute the given values into the formula to find the population
after 6 hours:
P(6) = 100 ·26
Step 3: Calculate the population after 6 hours:
P(6) = 100 ·26= 100 ·64 = 6400
Step 4: Therefore, after 6 hours, there will be 6400 bacteria in the popula-
tion.
Question 29
Question
A population of bacteria doubles every 3 hours. If there were initially 1000
bacteria in the population, how many bacteria will there be after 12 hours?
Round your answer to the nearest whole number.
Solution
Step 1: Let P(t)represent the population of bacteria after thours. Since the
population doubles every 3 hours, we can express this growth as an exponential
function:
P(t) = 1000 ·2t/3
Step 2: To find the population after 12 hours, substitute t= 12 into the
function:
P(12) = 1000 ·212/3= 1000 ·24= 1000 ·16 = 16000
Step 3: Therefore, after 12 hours, there will be approximately 16,000 bacteria
in the population.
Question 30
Question
Let f(x) = 32xand g(x) = log3(x). Find the value of f(g(9)).
20
Solution
Step 1: First, we need to find g(9).
g(9) = log3(9)
= 2 (Since 32= 9)
Step 2: Now, substitute g(9) = 2 into the function f(x)to find f(g(9)).
f(g(9)) = f(2)
= 32(2)
= 34
= 81
Therefore, f(g(9)) = 81.
Question 31
Question
Suppose a certain investment grows according to the formula A(t) = P(1 + r
n)nt,
where A(t)represents the amount of money in the account after tyears, Pis
the principal investment, ris the annual interest rate (expressed as a decimal),
and nis the number of times the interest is compounded per year. If P=$500,
r= 0.05, and n= 4, determine how long it will take for the investment to reach
$800.
Solution
Step 1: Substitute the given values into the formula A(t) = P(1 + r
n)nt to find
the equation that represents the amount of money in the account after tyears:
A(t) = 500 (1 + 0.05
4)4t
Step 2: Set up the equation 800 = 500 (1 + 0.05
4)4t, since we want to find
how long it will take for the investment to reach $800.
Step 3: Divide both sides of the equation by 500 to isolate the exponential
term:
800
500 =(1 + 0.05
4)4t
Step 4: Simplify the left side of the equation:
1.6 = (1 + 0.05
4)4t
21
Step 5: Rewrite the equation as an exponential equation:
1.6 = (1 + 0.0125)4t
Step 6: Simplify inside the parentheses:
1.6 = (1.0125)4t
Step 7: Take the natural logarithm of both sides to solve for t:
ln(1.6) = ln((1.0125)4t)
Step 8: Use the property of logarithms that allows us to bring down the
exponent:
ln(1.6) = 4tln(1.0125)
Step 9: Solve for tby dividing both sides by 4 ln(1.0125):
t=ln(1.6)
4 ln(1.0125)
Step 10: Use a calculator to approximate t:
t≈ln(1.6)
4 ln(1.0125) ≈0.4700
4(0.0124) ≈0.4700
0.0496 ≈9.48
Therefore, it will take approximately 9.48 years for the investment to reach
$800.
Question 32
Question
Solve for x:
52x−1=1
125
Solution
Step 1: Rewrite the fraction on the right side as a power of 5:
1
125 = 5−3
Step 2: Substitute 5−3on the right side of the equation:
52x−1= 5−3
Step 3: Since the bases are the same, we can equate the exponents:
2x−1 = −3
Step 4: Solve for x:
2x=−2
x=−1
Step 5: Therefore, the solution to the equation 52x−1=1
125 is x=−1.
22
Question 33
Question
Suppose a colony of bacteria triples in population every 6 hours. If the initial
population is 100 bacteria, what will be the population after 24 hours?
Solution
Step 1: Since the population triples every 6 hours, we can express the population
at any time tas P= 100 ·3t/6, where tis the time in hours.
Step 2: To find the population after 24 hours, we substitute t= 24 into the
equation:
P= 100 ·324/6= 100 ·34
Step 3: Calculating 34, we get:
34= 81
Step 4: Finally, we find the population after 24 hours:
P= 100 ·81 = 8100
Therefore, the population after 24 hours will be 8100 bacteria.
Question 34
Question
Solve the exponential equation 3x−2−2(3x) = 1.
Solution
Step 1: Let’s rewrite the equation by factoring out a 3xterm:
3x−2−2(3x) = 1
32·3−2−2(3x) = 1
3−2(32)−2(3x) = 1
1−2(3x) = 1
Step 2: Subtract 1 from both sides to get:
−2(3x) = 0
Step 3: Divide by −2to solve for 3x:
3x= 0
Step 4: Since 3xcannot be equal to 0 for any real value of x, this equation
has no solution.
23
Step 4: Multiply this value by 10,000 to find the amount of money in the
account after 5 years:
A≈10000 ×1.338225
A≈13382.25
Therefore, after 5 years, there will be approximately $13,382 in the account.
Question 2
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since the initial population count.
Determine the population of the city after 10 years, rounded to the nearest
whole number.
Solution
Step 1: Substitute t= 10 into the population function P(t)to find the popula-
tion after 10 years.
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210.
1.0210 = 1.218391]
Step 3: Substitute this value back into the population equation.
P(10) = 5000 ·1.218391
Step 4: Multiply to find the population after 10 years.
P(10) ≈6091
Therefore, the population of the city after 10 years is approximately 6091.
Question 3
Question
Samantha invests $5000 in a savings account that earns 3% interest compounded
quarterly. How much will the investment be worth after 10 years?
2
Solution
Let’s start by setting up the formula for compound interest, which is given by:
A=P(1 + r
n)nt
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). - nis the number of times that interest is
compounded per year. - tis the time the money is invested for in years.
In this case: - P= 5000 (the initial investment). - r= 0.03 (3% annual
interest rate as a decimal). - n= 4 (quarterly compounding). - t= 10 years.
Substitute these values into the compound interest formula to find out the
final amount after 10 years.
Step 1: Calculate the amount using the compound interest formula.
A= 5000 (1 + 0.03
4)4×10
Step 2: Simplify the expression inside the parentheses.
A= 5000 (1 + 0.0075)40
Step 3: Add the numbers inside the parentheses.
A= 5000 ×1.007540
Step 4: Calculate the value of 1.007540.
A≈5000 ×1.3498
Step 5: Multiply to find the final amount.
A≈6749.02
Therefore, the investment will be worth approximately $6749.02 after 10
years.
Question 4
Question
Suppose a population of bacteria doubles every 4 hours. If there are initially
100 bacteria, determine the exponential function that models the population of
bacteria after thours.
3
Solution
Step 1: Let P(t)be the population of bacteria after thours. Since the population
doubles every 4 hours, we can express P(t)as P(t) = 100 ·2(t/4).
Thus, the exponential function that models the population of bacteria after
thours is P(t) = 100 ·2(t/4).
Question 5
Question
A bacteria culture starts with 1000 bacteria and doubles in size every 3 hours.
Write an exponential growth model to represent the number of bacteria after t
hours, and determine how many bacteria there will be after 9 hours.
Solution
Step 1: To write the exponential growth model, we start with the general ex-
ponential function P(t) = P0·rt, where: - P(t)is the population after thours,
-P0is the initial population, - ris the growth rate per unit of time, and - tis
the time in hours.
Given that the bacteria doubles in size every 3 hours, the growth rate ris 2
(since doubling corresponds to multiplying by 2). The initial population P0is
1000. Therefore, the exponential growth model for this scenario is:
P(t) = 1000 ·2t/3
Step 2: To determine the number of bacteria after 9 hours, we substitute
t= 9 into the exponential growth model:
P(9) = 1000 ·29/3
P(9) = 1000 ·23
P(9) = 1000 ·8
P(9) = 8000
After 9 hours, there will be 8000 bacteria in the culture.
Question 6
Question
Suppose a population of insects is growing exponentially, with an initial popu-
lation of 100 insects and a growth rate of 20
4
Solution
Step 1: The exponential growth model for this scenario can be expressed as:
P(t) = P0·(1 + r)t
where: - P(t)is the population after tdays, - P0is the initial population, - ris
the growth rate per day (as a decimal), and - tis the number of days.
Step 2: Substituting the given values into the formula, we have:
P(t) = 100 ·(1 + 0.20)t
Step 3: Simplifying the expression, we get:
P(t) = 100 ·(1.20)t
P(t) = 100 ·1.20t
Therefore, the exponential growth model for this scenario is P(t) = 100 ·
1.20t.
Question 7
Question
Solve the following exponential equation: 3x+1 −3x= 12.
Solution
Step 1: We start by simplifying the left side of the equation using the properties
of exponents.
Step 2: Since 3x+1 = 3 ·3x, we rewrite the equation as 3·3x−3x= 12.
Step 3: Combining like terms, we get 2·3x= 12.
Step 4: Divide by 2 to isolate 3x:3x= 6.
Step 5: Now, we can rewrite 6as 31·2.
Step 6: By the properties of exponents, we can rewrite the equation as
3x= 31·3log32.
Step 7: Equating the exponents, we have x= 1 + log32.
Therefore, the solution to the equation 3x+1 −3x= 12 is x= 1 + log32.
Question 8
Question
Samantha invests $5000 in a savings account that pays an annual interest rate
of 4.5%, compounded continuously. How long will it take for her investment to
double in value?
5
Solution
Let A(t)be the amount of money in the account after tyears. We know that
the formula for continuously compounded interest is given by A(t) = P·ert,
where Pis the principal amount, ris the annual interest rate, and tis the time
in years.
Step 1: Substitute the given values into the formula. The principal amount
Pis $5000, the annual interest rate ris 0.045 (4.5% as a decimal), and we want
to find the time tit takes for the investment to double, so A(t) = 2P. Therefore,
the equation becomes:
2·5000 = 5000 ·e0.045t
Step 2: Solve for t. Dividing both sides by 5000, we get:
2 = e0.045t
Taking the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
Using the property ln(ex) = x:
ln(2) = 0.045t
Now, solving for t:
t=ln(2)
0.045 ≈0.6931
0.045 ≈15.47
Therefore, it will take about 15.47 years for Samantha’s investment to double
in value when compounded continuously.
Question 9
Question
Solve the exponential equation for x:2x−3−2x−1= 4.
Solution
To solve the given exponential equation, we need to manipulate the equation in
a way that allows us to isolate the variable x.
Step 1: Rewrite the equation using the properties of exponents.
2x−3−2x−1= 4.
Step 2: Recognize that 2x−3=1
2·2xand 2x−1= 2 ·2x.
1
2·2x−2·2x= 4.
6
Step 3: Combine like terms on the left side.
(1
2−2)·2x= 4.
Step 4: Simplify the left side of the equation.
−3
2·2x= 4.
Step 5: Solve for xby isolating 2x.
2x=−4
3
2
=−8
3.
Step 6: Rewrite the equation in exponential form.
x= log2(−8
3).
Since the logarithm of a negative number is undefined for real numbers, there
is no solution to the original exponential equation.
Question 10
Question
Suppose that an investment of $5000 grows according to the formula A= 5000 ·
e0.04t, where Ais the amount of money after tyears. How long will it take for
the investment to double?
Solution
Step 1: To find the time it takes for the investment to double, we need to solve
the equation 2·5000 = 5000 ·e0.04t.
Step 2: Simplifying the equation, we have 10000 = 5000 ·e0.04t.
Step 3: Dividing both sides by 5000, we get 2 = e0.04t.
Step 4: Taking the natural logarithm of both sides, we have ln(2) = ln(e0.04t).
Step 5: Using the property ln(ex) = x, we can simplify further to ln(2) =
0.04t.
Step 6: Next, we solve for tby dividing by 0.04:t=ln(2)
0.04 .
Step 7: Calculating this expression, t≈0.6931
0.04 ≈17.33 years.
Step 8: Therefore, it will take approximately 17.33 years for the investment
to double.
Question 11
Question
Solve the exponential equation 23x−1= 8.
7
Solution
Step 1: Rewrite 8as a power of 2.
23= 8
Step 2: Substitute 8as 23in the equation.
23x−1= 23
Step 3: Set the exponents equal to each other.
3x−1 = 3
Step 4: Solve for x.
3x= 4
x=4
3
Step 5: Check the solution by substituting x=4
3back into the original
equation.
23( 4
3)−1= 8
24−1= 8
23= 8
8 = 8
Therefore, the solution to the exponential equation is x=4
3.
Question 12
Question
Samantha invests $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much will Samantha have in the account after 10
years?
Solution
Let’s denote the amount of money in the account after tyears as A(t). We can
use the formula for continuous compounding:
A(t) = P·ert
where: - Pis the principal amount (initial investment), - ris the annual
interest rate (in decimal form), - tis the time in years, and - eis Euler’s number,
approximately equal to 2.71828.
8
In this case, P= $5000,r= 0.035, and t= 10. Plugging these values into
the formula, we get:
Step 1:
A(t) = 5000 ·e0.035·10
Step 2:
A(10) = 5000 ·e0.35
Step 3:
A(10) ≈5000 ·1.419067
Step 4:
A(10) ≈7095.335
Therefore, Samantha will have approximately $7095.34 in the account after
10 years.
Question 13
Question
Solve for xin the equation log2(2x+ 5) + log2(x−1) = 3.
Solution
Step 1: Rewrite the equation using the properties of logarithms.
log2(2x+ 5) + log2(x−1) = 3
log2((2x+ 5)(x−1)) = 3
Step 2: Remove the logarithm by expressing the equation in exponential
form.
23= (2x+ 5)(x−1)
8 = 2x2+ 3x−5
Step 3: Rearrange the equation into standard form by setting it equal to
zero.
2x2+ 3x−13 = 0
Step 4: Solve the quadratic equation using the quadratic formula x=
−b±√b2−4ac
2a. Using the values a= 2,b= 3, and c=−13:
x=−3±√32−4(2)(−13)
2(2)
x=−3±√9 + 104
4
x=−3±√113
4
Therefore, the solutions for xare x=−3+√113
4and x=−3−√113
4.
9
Question 14
Question
Samantha invests $10,000 in an account that pays 4.5% interest compounded
continuously. How long will it take for her investment to double?
Solution
Step 1: To find the time it takes for Samantha’s investment to double, we need
to use the formula for continuously compounded interest:
A=P·ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= the interest rate (decimal form), and t= the
time the money is invested for.
Step 2: Since Samantha wants to double her investment, we can set up the
equation as follows:
2·P=P·ert
Step 3: Substitute the given values into the equation:
2·10,000 = 10,000 ·e0.045t
Step 4: Simplify the equation:
20,000 = 10,000 ·e0.045t
Step 5: Divide both sides by 10,000 to solve for e0.045t:
2 = e0.045t
Step 6: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
Step 7: Remember that ln(ex) = x, so we have:
ln(2) = 0.045t
Step 8: Solve for tby dividing by 0.045:
t=ln(2)
0.045
Step 9: Using a calculator to approximate the value:
t≈0.6931
0.045 ≈15.47 years
Answer: It will take approximately 15.47 years for her investment to double
when earning 4.5% interest compounded continuously.
10
Question 15
Question
Suppose an investment account starts with an initial deposit of $5000 and earns
3% interest compounded annually. How long will it take for the account balance
to reach $8000?
Solution
Step 1: First, we need to determine the exponential growth model for this
scenario. The formula for compound interest is given by:
A=P(1 + r)t
where: - Ais the final amount in the account - Pis the principal amount (initial
deposit) - ris the annual interest rate in decimal form - tis the time the money
is invested for in years
Step 2: Substituting the given values into the formula:
8000 = 5000(1 + 0.03)t
Step 3: Divide both sides by 5000 to isolate the exponential term:
8000
5000 = (1 + 0.03)t
1.6 = (1.03)t
Step 4: To solve for t, we will take the natural logarithm of both sides to
eliminate the exponent:
ln(1.6) = ln(1.03)t
Step 5: By the properties of logarithms, we can bring down the exponent t:
ln(1.6) = t·ln(1.03)
Step 6: Finally, we solve for tby dividing both sides by ln(1.03):
t=ln(1.6)
ln(1.03) ≈0.4700
0.0296 ≈15.88
Therefore, it will take approximately 15.88 years for the account balance to
reach $8000.
Question 16
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since 2020. Determine the approximate year
when the population of the city will reach 10,000 people.
11
Solution
Step 1: Let’s set up the equation with the given information:
10,000 = 5000 ·1.02t
Step 2: Divide both sides by 5000 to isolate the exponential term:
2 = 1.02t
Step 3: To solve for t, we need to express the equation in logarithmic form:
log1.02 2 = t
Step 4: Use a calculator to compute log1.02 2:
t≈34.3
Step 5: This means that it will take approximately 34.3 years since 2020 for
the population to reach 10,000 people.
Step 6: Finally, to determine the year when this population will be reached,
add 34.3 to 2020:
2020 + 34.3≈2054.3
Therefore, the approximate year when the population of the city will reach
10,000 people is 2054.
Question 17
Question
Solve the following exponential equation for x:
4x+1 = 32
Solution
Step 1: Rewrite 32 as a power of 4. Step 2: Solve for xusing the properties of
exponents.
Step 1: We can rewrite 32 as a power of 4 by noticing that 32 = 42·2.
Therefore, 32 = 42·41= 42+1.
Step 2: Substitute 32 with 42+1 in the equation 4x+1 = 32:
4x+1 = 42+1
By the property of equality for exponents, we can equate the exponents:
x+ 1 = 2 + 1
Solving for xgives:
x= 2
Therefore, the solution to the equation 4x+1 = 32 is x= 2.
12
Question 18
Question
Solve the equation for x:32x= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 33= 27,we can rewrite 27 as 33.
Step 2: Rewrite the equation 32x= 27 with 27 expressed as 33.
32x= 33.
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x= 3.
Step 4: Solve for x.
2x= 3
x=3
2
x= 1.5.
Therefore, the solution to the equation 32x= 27 is x= 1.5.
Question 19
Question
Samantha invested $5000 in an account that earns 3% annual interest com-
pounded continuously. How long will it take for the investment to double in
value?
Solution
Step 1: The continuous compound interest formula is given by A=P ert, where:
-Ais the amount of money after tyears, - Pis the principal investment amount,
-ris the annual interest rate (as a decimal), and - tis the time in years.
Step 2: Since we want to know when the investment will double, we can set
up the equation 2P=P e0.03t, where P= 5000 and r= 0.03.
Step 3: Simplifying the equation, we get 2 = e0.03t.
Step 4: To solve for t, we take the natural logarithm of both sides: ln(2) =
ln(e0.03t).
Step 5: Using the properties of logarithms, we have ln(2) = 0.03t.
13
Step 6: Divide by 0.03 to solve for t:t=ln(2)
0.03 .
Step 7: Calculating the value of t, we have t≈ln(2)
0.03 ≈0.693
0.03 ≈23.1.
Step 8: Therefore, it will take approximately 23.1years for Samantha’s
investment to double in value.
Question 20
Question
Solve for xin the equation 23x−5 = 13.
Solution
Step 1: Add 5 to both sides of the equation to isolate the exponential term:
23x= 13 + 5 = 18
Step 2: Rewrite 18 as a power of 2:
23x= 2log218
Step 3: Set the exponents equal to each other:
3x= log218
Step 4: Solve for xby dividing both sides by 3:
x=log218
3
Step 5: Simplify the expression for xusing properties of logarithms:
x=log2(2log218)
3
x=log218
3
Therefore, the solution to the equation 23x−5 = 13 is x=log218
3.
Question 21
Question
Solve the following exponential equation for x:32x−3x+1 + 2 = 0.
14
Solution
Step 1: Let’s rewrite the equation in terms of 3x:
32x−3x+1 + 2 = 0
(3x)2−3·3x+ 2 = 0
Step 2: Now, let’s substitute y= 3xto simplify the equation:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
Step 4: Set each factor to zero and solve for y:
y−1 = 0 =⇒y= 1
y−2 = 0 =⇒y= 2
Step 5: Recall that y= 3x: For y= 1:
3x= 1
x= 0
For y= 2:
3x= 2
x= log32
Step 6: Therefore, the solutions to the equation are x= 0 and x= log32.
Question 22
Question
A certain population of bacteria doubles every 5 hours. If the initial population
is 1000 bacteria, find an exponential model expressing the population Pas a
function of time tin hours. Additionally, determine the population after 15
hours.
Solution
Step 1: To find the exponential model, we use the formula for exponential
growth: P(t) = P0·2(t/k), where P0is the initial population, kis the time it
takes for the population to double, and tis the time elapsed.
Step 2: Given that the population doubles every 5 hours, we have k= 5.
Also, the initial population P0= 1000. Therefore, the exponential model is
P(t) = 1000 ·2(t/5).
Step 3: To find the population after 15 hours, we substitute t= 15 into the
model: P(15) = 1000 ·2(15/5).
Step 4: Simplifying the expression gives P(15) = 1000 ·23= 1000 ·8 = 8000.
Therefore, the population after 15 hours is 8000 bacteria.
15
Question 23
Question
Suppose a certain investment grows according to the formula A(t) = 5000·e0.04t,
where A(t)represents the amount of money in the investment after tyears. Find
the rate of change of the investment after 4 years.
Solution
Step 1: Calculate the derivative of A(t)with respect to t.
A′(t) = d
dt (5000 ·e0.04t)
Step 2: Apply the chain rule to find the derivative of A(t).
A′(t) = 5000 ·d
dt (e0.04t)
Step 3: Differentiate e0.04twith respect to t.
A′(t) = 5000 ·0.04 ·e0.04t
A′(t) = 200e0.04t
Step 4: Evaluate the rate of change after 4 years by plugging in t= 4.
A′(4) = 200e0.04·4
A′(4) = 200e0.16
A′(4) ≈226.90
Therefore, the rate of change of the investment after 4 years is approximately
226.90.
Question 24
Question
Samantha deposits $5000 into a savings account that earns an annual interest
rate of 3%. How long will it take for the account balance to double if the interest
is compounded continuously?
16
Solution
Step 1: Let A(t)represent the amount of money in the savings account after t
years. We can use the formula for continuously compounded interest:
A(t) = P·ert,
where Pis the principal amount ($5000 in this case), ris the annual interest
rate (expressed as a decimal, so 0.03 for 3%), and tis the time in years.
Step 2: Since we want to find when the account balance doubles, we are
looking for tsuch that A(t) = 2 ·P= 2 ·$5000 = $10000. Substituting this into
the formula, we get:
10000 = 5000 ·e0.03t.
Step 3: Divide both sides by 5000 to isolate the exponential term:
10000
5000 =e0.03t.
Step 4: Simplify the left side to get:
2 = e0.03t.
Step 5: To solve for t, take the natural logarithm of both sides:
ln(2) = ln(e0.03t).
Step 6: Using the property ln(ex) = x, we get:
ln(2) = 0.03t.
Step 7: Finally, solve for tby dividing by 0.03:
t=ln(2)
0.03 ≈23.1.
Therefore, it will take approximately 23.1 years for the account balance to
double with continuously compounded interest.
Question 25
Question
Let f(x) = 3xand g(x) = log3(x). Determine the domain of the composite
function g(f(x)).
17
Solution
Step 1: Find the composite function g(f(x)):
g(f(x)) = log3(3x)
Step 2: Use the property of logarithmic functions loga(ax) = x:
g(f(x)) = x
Step 3: Determine the domain of g(f(x)). Since xis a real number, the
domain of g(f(x)) is also all real numbers. Therefore, the domain of g(f(x)) is
(−∞,∞).
Question 26
Question
Solve the exponential equation 32x−1= 81 for x.
Solution
Step 1: Rewrite 81 as a power of 3:81 = 34.
Step 2: Substitute 81 with 34in the equation: 32x−1= 34.
Step 3: Since the bases on both sides are the same, we can set the exponents
equal to each other: 2x−1 = 4.
Step 4: Solve the resulting linear equation for x:
2x−1 = 4
2x= 5
x=5
2
Step 5: Therefore, the solution to the exponential equation 32x−1= 81 is
x=5
2.
Question 27
Question
Samantha invests $10,000 in an account that pays 3.5% interest compounded
continuously. How long will it take for her investment to double in value?
18
Solution
Step 1: First, we need to find the formula for continuously compounded interest.
The formula is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the princi-
pal amount (initial investment), - ris the annual interest rate (expressed as a
decimal), - tis the time the money is invested for, - eis the base of the natural
logarithm, approximately equal to 2.71828.
Step 2: Since Samantha wants her investment to double, we set A= 2Pin
the formula:
2P=P·e0.035t
Step 3: Divide both sides by Pto simplify the equation:
2 = e0.035t
Step 4: Take the natural logarithm of both sides to solve for t:
ln 2 = ln e0.035t
ln 2 = 0.035t·ln e
ln 2 = 0.035t
Step 5: Solve for tby dividing both sides by 0.035:
t=ln 2
0.035
t≈0.6931
0.035
t≈19.802 years
Therefore, it will take approximately 19.802 years for Samantha’s investment
to double in value when compounded continuously at a rate of 3.5%.
Question 28
Question
A population of bacteria doubles every hour. If there are initially 100 bacteria,
how many bacteria will there be after 6 hours?
19
Solution
Step 1: Let’s denote the initial number of bacteria as P0= 100 and the growth
factor per hour as r= 2, since the population doubles every hour. The popula-
tion after thours can be represented by the exponential function P(t) = P0·rt.
Step 2: Substitute the given values into the formula to find the population
after 6 hours:
P(6) = 100 ·26
Step 3: Calculate the population after 6 hours:
P(6) = 100 ·26= 100 ·64 = 6400
Step 4: Therefore, after 6 hours, there will be 6400 bacteria in the popula-
tion.
Question 29
Question
A population of bacteria doubles every 3 hours. If there were initially 1000
bacteria in the population, how many bacteria will there be after 12 hours?
Round your answer to the nearest whole number.
Solution
Step 1: Let P(t)represent the population of bacteria after thours. Since the
population doubles every 3 hours, we can express this growth as an exponential
function:
P(t) = 1000 ·2t/3
Step 2: To find the population after 12 hours, substitute t= 12 into the
function:
P(12) = 1000 ·212/3= 1000 ·24= 1000 ·16 = 16000
Step 3: Therefore, after 12 hours, there will be approximately 16,000 bacteria
in the population.
Question 30
Question
Let f(x) = 32xand g(x) = log3(x). Find the value of f(g(9)).
20
Solution
Step 1: First, we need to find g(9).
g(9) = log3(9)
= 2 (Since 32= 9)
Step 2: Now, substitute g(9) = 2 into the function f(x)to find f(g(9)).
f(g(9)) = f(2)
= 32(2)
= 34
= 81
Therefore, f(g(9)) = 81.
Question 31
Question
Suppose a certain investment grows according to the formula A(t) = P(1 + r
n)nt,
where A(t)represents the amount of money in the account after tyears, Pis
the principal investment, ris the annual interest rate (expressed as a decimal),
and nis the number of times the interest is compounded per year. If P=$500,
r= 0.05, and n= 4, determine how long it will take for the investment to reach
$800.
Solution
Step 1: Substitute the given values into the formula A(t) = P(1 + r
n)nt to find
the equation that represents the amount of money in the account after tyears:
A(t) = 500 (1 + 0.05
4)4t
Step 2: Set up the equation 800 = 500 (1 + 0.05
4)4t, since we want to find
how long it will take for the investment to reach $800.
Step 3: Divide both sides of the equation by 500 to isolate the exponential
term:
800
500 =(1 + 0.05
4)4t
Step 4: Simplify the left side of the equation:
1.6 = (1 + 0.05
4)4t
21
Step 5: Rewrite the equation as an exponential equation:
1.6 = (1 + 0.0125)4t
Step 6: Simplify inside the parentheses:
1.6 = (1.0125)4t
Step 7: Take the natural logarithm of both sides to solve for t:
ln(1.6) = ln((1.0125)4t)
Step 8: Use the property of logarithms that allows us to bring down the
exponent:
ln(1.6) = 4tln(1.0125)
Step 9: Solve for tby dividing both sides by 4 ln(1.0125):
t=ln(1.6)
4 ln(1.0125)
Step 10: Use a calculator to approximate t:
t≈ln(1.6)
4 ln(1.0125) ≈0.4700
4(0.0124) ≈0.4700
0.0496 ≈9.48
Therefore, it will take approximately 9.48 years for the investment to reach
$800.
Question 32
Question
Solve for x:
52x−1=1
125
Solution
Step 1: Rewrite the fraction on the right side as a power of 5:
1
125 = 5−3
Step 2: Substitute 5−3on the right side of the equation:
52x−1= 5−3
Step 3: Since the bases are the same, we can equate the exponents:
2x−1 = −3
Step 4: Solve for x:
2x=−2
x=−1
Step 5: Therefore, the solution to the equation 52x−1=1
125 is x=−1.
22
Question 33
Question
Suppose a colony of bacteria triples in population every 6 hours. If the initial
population is 100 bacteria, what will be the population after 24 hours?
Solution
Step 1: Since the population triples every 6 hours, we can express the population
at any time tas P= 100 ·3t/6, where tis the time in hours.
Step 2: To find the population after 24 hours, we substitute t= 24 into the
equation:
P= 100 ·324/6= 100 ·34
Step 3: Calculating 34, we get:
34= 81
Step 4: Finally, we find the population after 24 hours:
P= 100 ·81 = 8100
Therefore, the population after 24 hours will be 8100 bacteria.
Question 34
Question
Solve the exponential equation 3x−2−2(3x) = 1.
Solution
Step 1: Let’s rewrite the equation by factoring out a 3xterm:
3x−2−2(3x) = 1
32·3−2−2(3x) = 1
3−2(32)−2(3x) = 1
1−2(3x) = 1
Step 2: Subtract 1 from both sides to get:
−2(3x) = 0
Step 3: Divide by −2to solve for 3x:
3x= 0
Step 4: Since 3xcannot be equal to 0 for any real value of x, this equation
has no solution.
23
Step 4: Multiply this value by 10,000 to find the amount of money in the
account after 5 years:
A≈10000 ×1.338225
A≈13382.25
Therefore, after 5 years, there will be approximately $13,382 in the account.
Question 2
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since the initial population count.
Determine the population of the city after 10 years, rounded to the nearest
whole number.
Solution
Step 1: Substitute t= 10 into the population function P(t)to find the popula-
tion after 10 years.
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210.
1.0210 = 1.218391]
Step 3: Substitute this value back into the population equation.
P(10) = 5000 ·1.218391
Step 4: Multiply to find the population after 10 years.
P(10) ≈6091
Therefore, the population of the city after 10 years is approximately 6091.
Question 3
Question
Samantha invests $5000 in a savings account that earns 3% interest compounded
quarterly. How much will the investment be worth after 10 years?
2
Solution
Let’s start by setting up the formula for compound interest, which is given by:
A=P(1 + r
n)nt
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). - nis the number of times that interest is
compounded per year. - tis the time the money is invested for in years.
In this case: - P= 5000 (the initial investment). - r= 0.03 (3% annual
interest rate as a decimal). - n= 4 (quarterly compounding). - t= 10 years.
Substitute these values into the compound interest formula to find out the
final amount after 10 years.
Step 1: Calculate the amount using the compound interest formula.
A= 5000 (1 + 0.03
4)4×10
Step 2: Simplify the expression inside the parentheses.
A= 5000 (1 + 0.0075)40
Step 3: Add the numbers inside the parentheses.
A= 5000 ×1.007540
Step 4: Calculate the value of 1.007540.
A≈5000 ×1.3498
Step 5: Multiply to find the final amount.
A≈6749.02
Therefore, the investment will be worth approximately $6749.02 after 10
years.
Question 4
Question
Suppose a population of bacteria doubles every 4 hours. If there are initially
100 bacteria, determine the exponential function that models the population of
bacteria after thours.
3
Solution
Step 1: Let P(t)be the population of bacteria after thours. Since the population
doubles every 4 hours, we can express P(t)as P(t) = 100 ·2(t/4).
Thus, the exponential function that models the population of bacteria after
thours is P(t) = 100 ·2(t/4).
Question 5
Question
A bacteria culture starts with 1000 bacteria and doubles in size every 3 hours.
Write an exponential growth model to represent the number of bacteria after t
hours, and determine how many bacteria there will be after 9 hours.
Solution
Step 1: To write the exponential growth model, we start with the general ex-
ponential function P(t) = P0·rt, where: - P(t)is the population after thours,
-P0is the initial population, - ris the growth rate per unit of time, and - tis
the time in hours.
Given that the bacteria doubles in size every 3 hours, the growth rate ris 2
(since doubling corresponds to multiplying by 2). The initial population P0is
1000. Therefore, the exponential growth model for this scenario is:
P(t) = 1000 ·2t/3
Step 2: To determine the number of bacteria after 9 hours, we substitute
t= 9 into the exponential growth model:
P(9) = 1000 ·29/3
P(9) = 1000 ·23
P(9) = 1000 ·8
P(9) = 8000
After 9 hours, there will be 8000 bacteria in the culture.
Question 6
Question
Suppose a population of insects is growing exponentially, with an initial popu-
lation of 100 insects and a growth rate of 20
4
Solution
Step 1: The exponential growth model for this scenario can be expressed as:
P(t) = P0·(1 + r)t
where: - P(t)is the population after tdays, - P0is the initial population, - ris
the growth rate per day (as a decimal), and - tis the number of days.
Step 2: Substituting the given values into the formula, we have:
P(t) = 100 ·(1 + 0.20)t
Step 3: Simplifying the expression, we get:
P(t) = 100 ·(1.20)t
P(t) = 100 ·1.20t
Therefore, the exponential growth model for this scenario is P(t) = 100 ·
1.20t.
Question 7
Question
Solve the following exponential equation: 3x+1 −3x= 12.
Solution
Step 1: We start by simplifying the left side of the equation using the properties
of exponents.
Step 2: Since 3x+1 = 3 ·3x, we rewrite the equation as 3·3x−3x= 12.
Step 3: Combining like terms, we get 2·3x= 12.
Step 4: Divide by 2 to isolate 3x:3x= 6.
Step 5: Now, we can rewrite 6as 31·2.
Step 6: By the properties of exponents, we can rewrite the equation as
3x= 31·3log32.
Step 7: Equating the exponents, we have x= 1 + log32.
Therefore, the solution to the equation 3x+1 −3x= 12 is x= 1 + log32.
Question 8
Question
Samantha invests $5000 in a savings account that pays an annual interest rate
of 4.5%, compounded continuously. How long will it take for her investment to
double in value?
5
Solution
Let A(t)be the amount of money in the account after tyears. We know that
the formula for continuously compounded interest is given by A(t) = P·ert,
where Pis the principal amount, ris the annual interest rate, and tis the time
in years.
Step 1: Substitute the given values into the formula. The principal amount
Pis $5000, the annual interest rate ris 0.045 (4.5% as a decimal), and we want
to find the time tit takes for the investment to double, so A(t) = 2P. Therefore,
the equation becomes:
2·5000 = 5000 ·e0.045t
Step 2: Solve for t. Dividing both sides by 5000, we get:
2 = e0.045t
Taking the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
Using the property ln(ex) = x:
ln(2) = 0.045t
Now, solving for t:
t=ln(2)
0.045 ≈0.6931
0.045 ≈15.47
Therefore, it will take about 15.47 years for Samantha’s investment to double
in value when compounded continuously.
Question 9
Question
Solve the exponential equation for x:2x−3−2x−1= 4.
Solution
To solve the given exponential equation, we need to manipulate the equation in
a way that allows us to isolate the variable x.
Step 1: Rewrite the equation using the properties of exponents.
2x−3−2x−1= 4.
Step 2: Recognize that 2x−3=1
2·2xand 2x−1= 2 ·2x.
1
2·2x−2·2x= 4.
6
Step 3: Combine like terms on the left side.
(1
2−2)·2x= 4.
Step 4: Simplify the left side of the equation.
−3
2·2x= 4.
Step 5: Solve for xby isolating 2x.
2x=−4
3
2
=−8
3.
Step 6: Rewrite the equation in exponential form.
x= log2(−8
3).
Since the logarithm of a negative number is undefined for real numbers, there
is no solution to the original exponential equation.
Question 10
Question
Suppose that an investment of $5000 grows according to the formula A= 5000 ·
e0.04t, where Ais the amount of money after tyears. How long will it take for
the investment to double?
Solution
Step 1: To find the time it takes for the investment to double, we need to solve
the equation 2·5000 = 5000 ·e0.04t.
Step 2: Simplifying the equation, we have 10000 = 5000 ·e0.04t.
Step 3: Dividing both sides by 5000, we get 2 = e0.04t.
Step 4: Taking the natural logarithm of both sides, we have ln(2) = ln(e0.04t).
Step 5: Using the property ln(ex) = x, we can simplify further to ln(2) =
0.04t.
Step 6: Next, we solve for tby dividing by 0.04:t=ln(2)
0.04 .
Step 7: Calculating this expression, t≈0.6931
0.04 ≈17.33 years.
Step 8: Therefore, it will take approximately 17.33 years for the investment
to double.
Question 11
Question
Solve the exponential equation 23x−1= 8.
7
Solution
Step 1: Rewrite 8as a power of 2.
23= 8
Step 2: Substitute 8as 23in the equation.
23x−1= 23
Step 3: Set the exponents equal to each other.
3x−1 = 3
Step 4: Solve for x.
3x= 4
x=4
3
Step 5: Check the solution by substituting x=4
3back into the original
equation.
23( 4
3)−1= 8
24−1= 8
23= 8
8 = 8
Therefore, the solution to the exponential equation is x=4
3.
Question 12
Question
Samantha invests $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much will Samantha have in the account after 10
years?
Solution
Let’s denote the amount of money in the account after tyears as A(t). We can
use the formula for continuous compounding:
A(t) = P·ert
where: - Pis the principal amount (initial investment), - ris the annual
interest rate (in decimal form), - tis the time in years, and - eis Euler’s number,
approximately equal to 2.71828.
8
In this case, P= $5000,r= 0.035, and t= 10. Plugging these values into
the formula, we get:
Step 1:
A(t) = 5000 ·e0.035·10
Step 2:
A(10) = 5000 ·e0.35
Step 3:
A(10) ≈5000 ·1.419067
Step 4:
A(10) ≈7095.335
Therefore, Samantha will have approximately $7095.34 in the account after
10 years.
Question 13
Question
Solve for xin the equation log2(2x+ 5) + log2(x−1) = 3.
Solution
Step 1: Rewrite the equation using the properties of logarithms.
log2(2x+ 5) + log2(x−1) = 3
log2((2x+ 5)(x−1)) = 3
Step 2: Remove the logarithm by expressing the equation in exponential
form.
23= (2x+ 5)(x−1)
8 = 2x2+ 3x−5
Step 3: Rearrange the equation into standard form by setting it equal to
zero.
2x2+ 3x−13 = 0
Step 4: Solve the quadratic equation using the quadratic formula x=
−b±√b2−4ac
2a. Using the values a= 2,b= 3, and c=−13:
x=−3±√32−4(2)(−13)
2(2)
x=−3±√9 + 104
4
x=−3±√113
4
Therefore, the solutions for xare x=−3+√113
4and x=−3−√113
4.
9
Question 14
Question
Samantha invests $10,000 in an account that pays 4.5% interest compounded
continuously. How long will it take for her investment to double?
Solution
Step 1: To find the time it takes for Samantha’s investment to double, we need
to use the formula for continuously compounded interest:
A=P·ert
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= the interest rate (decimal form), and t= the
time the money is invested for.
Step 2: Since Samantha wants to double her investment, we can set up the
equation as follows:
2·P=P·ert
Step 3: Substitute the given values into the equation:
2·10,000 = 10,000 ·e0.045t
Step 4: Simplify the equation:
20,000 = 10,000 ·e0.045t
Step 5: Divide both sides by 10,000 to solve for e0.045t:
2 = e0.045t
Step 6: Take the natural logarithm of both sides to solve for t:
ln(2) = ln(e0.045t)
Step 7: Remember that ln(ex) = x, so we have:
ln(2) = 0.045t
Step 8: Solve for tby dividing by 0.045:
t=ln(2)
0.045
Step 9: Using a calculator to approximate the value:
t≈0.6931
0.045 ≈15.47 years
Answer: It will take approximately 15.47 years for her investment to double
when earning 4.5% interest compounded continuously.
10
Question 15
Question
Suppose an investment account starts with an initial deposit of $5000 and earns
3% interest compounded annually. How long will it take for the account balance
to reach $8000?
Solution
Step 1: First, we need to determine the exponential growth model for this
scenario. The formula for compound interest is given by:
A=P(1 + r)t
where: - Ais the final amount in the account - Pis the principal amount (initial
deposit) - ris the annual interest rate in decimal form - tis the time the money
is invested for in years
Step 2: Substituting the given values into the formula:
8000 = 5000(1 + 0.03)t
Step 3: Divide both sides by 5000 to isolate the exponential term:
8000
5000 = (1 + 0.03)t
1.6 = (1.03)t
Step 4: To solve for t, we will take the natural logarithm of both sides to
eliminate the exponent:
ln(1.6) = ln(1.03)t
Step 5: By the properties of logarithms, we can bring down the exponent t:
ln(1.6) = t·ln(1.03)
Step 6: Finally, we solve for tby dividing both sides by ln(1.03):
t=ln(1.6)
ln(1.03) ≈0.4700
0.0296 ≈15.88
Therefore, it will take approximately 15.88 years for the account balance to
reach $8000.
Question 16
Question
The population of a city is modeled by the function P(t) = 5000 ·1.02t, where
trepresents the number of years since 2020. Determine the approximate year
when the population of the city will reach 10,000 people.
11
Solution
Step 1: Let’s set up the equation with the given information:
10,000 = 5000 ·1.02t
Step 2: Divide both sides by 5000 to isolate the exponential term:
2 = 1.02t
Step 3: To solve for t, we need to express the equation in logarithmic form:
log1.02 2 = t
Step 4: Use a calculator to compute log1.02 2:
t≈34.3
Step 5: This means that it will take approximately 34.3 years since 2020 for
the population to reach 10,000 people.
Step 6: Finally, to determine the year when this population will be reached,
add 34.3 to 2020:
2020 + 34.3≈2054.3
Therefore, the approximate year when the population of the city will reach
10,000 people is 2054.
Question 17
Question
Solve the following exponential equation for x:
4x+1 = 32
Solution
Step 1: Rewrite 32 as a power of 4. Step 2: Solve for xusing the properties of
exponents.
Step 1: We can rewrite 32 as a power of 4 by noticing that 32 = 42·2.
Therefore, 32 = 42·41= 42+1.
Step 2: Substitute 32 with 42+1 in the equation 4x+1 = 32:
4x+1 = 42+1
By the property of equality for exponents, we can equate the exponents:
x+ 1 = 2 + 1
Solving for xgives:
x= 2
Therefore, the solution to the equation 4x+1 = 32 is x= 2.
12
Question 18
Question
Solve the equation for x:32x= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 33= 27,we can rewrite 27 as 33.
Step 2: Rewrite the equation 32x= 27 with 27 expressed as 33.
32x= 33.
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x= 3.
Step 4: Solve for x.
2x= 3
x=3
2
x= 1.5.
Therefore, the solution to the equation 32x= 27 is x= 1.5.
Question 19
Question
Samantha invested $5000 in an account that earns 3% annual interest com-
pounded continuously. How long will it take for the investment to double in
value?
Solution
Step 1: The continuous compound interest formula is given by A=P ert, where:
-Ais the amount of money after tyears, - Pis the principal investment amount,
-ris the annual interest rate (as a decimal), and - tis the time in years.
Step 2: Since we want to know when the investment will double, we can set
up the equation 2P=P e0.03t, where P= 5000 and r= 0.03.
Step 3: Simplifying the equation, we get 2 = e0.03t.
Step 4: To solve for t, we take the natural logarithm of both sides: ln(2) =
ln(e0.03t).
Step 5: Using the properties of logarithms, we have ln(2) = 0.03t.
13
Step 6: Divide by 0.03 to solve for t:t=ln(2)
0.03 .
Step 7: Calculating the value of t, we have t≈ln(2)
0.03 ≈0.693
0.03 ≈23.1.
Step 8: Therefore, it will take approximately 23.1years for Samantha’s
investment to double in value.
Question 20
Question
Solve for xin the equation 23x−5 = 13.
Solution
Step 1: Add 5 to both sides of the equation to isolate the exponential term:
23x= 13 + 5 = 18
Step 2: Rewrite 18 as a power of 2:
23x= 2log218
Step 3: Set the exponents equal to each other:
3x= log218
Step 4: Solve for xby dividing both sides by 3:
x=log218
3
Step 5: Simplify the expression for xusing properties of logarithms:
x=log2(2log218)
3
x=log218
3
Therefore, the solution to the equation 23x−5 = 13 is x=log218
3.
Question 21
Question
Solve the following exponential equation for x:32x−3x+1 + 2 = 0.
14
Solution
Step 1: Let’s rewrite the equation in terms of 3x:
32x−3x+1 + 2 = 0
(3x)2−3·3x+ 2 = 0
Step 2: Now, let’s substitute y= 3xto simplify the equation:
y2−3y+ 2 = 0
Step 3: Factor the quadratic equation:
(y−1)(y−2) = 0
Step 4: Set each factor to zero and solve for y:
y−1 = 0 =⇒y= 1
y−2 = 0 =⇒y= 2
Step 5: Recall that y= 3x: For y= 1:
3x= 1
x= 0
For y= 2:
3x= 2
x= log32
Step 6: Therefore, the solutions to the equation are x= 0 and x= log32.
Question 22
Question
A certain population of bacteria doubles every 5 hours. If the initial population
is 1000 bacteria, find an exponential model expressing the population Pas a
function of time tin hours. Additionally, determine the population after 15
hours.
Solution
Step 1: To find the exponential model, we use the formula for exponential
growth: P(t) = P0·2(t/k), where P0is the initial population, kis the time it
takes for the population to double, and tis the time elapsed.
Step 2: Given that the population doubles every 5 hours, we have k= 5.
Also, the initial population P0= 1000. Therefore, the exponential model is
P(t) = 1000 ·2(t/5).
Step 3: To find the population after 15 hours, we substitute t= 15 into the
model: P(15) = 1000 ·2(15/5).
Step 4: Simplifying the expression gives P(15) = 1000 ·23= 1000 ·8 = 8000.
Therefore, the population after 15 hours is 8000 bacteria.
15
Question 23
Question
Suppose a certain investment grows according to the formula A(t) = 5000·e0.04t,
where A(t)represents the amount of money in the investment after tyears. Find
the rate of change of the investment after 4 years.
Solution
Step 1: Calculate the derivative of A(t)with respect to t.
A′(t) = d
dt (5000 ·e0.04t)
Step 2: Apply the chain rule to find the derivative of A(t).
A′(t) = 5000 ·d
dt (e0.04t)
Step 3: Differentiate e0.04twith respect to t.
A′(t) = 5000 ·0.04 ·e0.04t
A′(t) = 200e0.04t
Step 4: Evaluate the rate of change after 4 years by plugging in t= 4.
A′(4) = 200e0.04·4
A′(4) = 200e0.16
A′(4) ≈226.90
Therefore, the rate of change of the investment after 4 years is approximately
226.90.
Question 24
Question
Samantha deposits $5000 into a savings account that earns an annual interest
rate of 3%. How long will it take for the account balance to double if the interest
is compounded continuously?
16
Solution
Step 1: Let A(t)represent the amount of money in the savings account after t
years. We can use the formula for continuously compounded interest:
A(t) = P·ert,
where Pis the principal amount ($5000 in this case), ris the annual interest
rate (expressed as a decimal, so 0.03 for 3%), and tis the time in years.
Step 2: Since we want to find when the account balance doubles, we are
looking for tsuch that A(t) = 2 ·P= 2 ·$5000 = $10000. Substituting this into
the formula, we get:
10000 = 5000 ·e0.03t.
Step 3: Divide both sides by 5000 to isolate the exponential term:
10000
5000 =e0.03t.
Step 4: Simplify the left side to get:
2 = e0.03t.
Step 5: To solve for t, take the natural logarithm of both sides:
ln(2) = ln(e0.03t).
Step 6: Using the property ln(ex) = x, we get:
ln(2) = 0.03t.
Step 7: Finally, solve for tby dividing by 0.03:
t=ln(2)
0.03 ≈23.1.
Therefore, it will take approximately 23.1 years for the account balance to
double with continuously compounded interest.
Question 25
Question
Let f(x) = 3xand g(x) = log3(x). Determine the domain of the composite
function g(f(x)).
17
Solution
Step 1: Find the composite function g(f(x)):
g(f(x)) = log3(3x)
Step 2: Use the property of logarithmic functions loga(ax) = x:
g(f(x)) = x
Step 3: Determine the domain of g(f(x)). Since xis a real number, the
domain of g(f(x)) is also all real numbers. Therefore, the domain of g(f(x)) is
(−∞,∞).
Question 26
Question
Solve the exponential equation 32x−1= 81 for x.
Solution
Step 1: Rewrite 81 as a power of 3:81 = 34.
Step 2: Substitute 81 with 34in the equation: 32x−1= 34.
Step 3: Since the bases on both sides are the same, we can set the exponents
equal to each other: 2x−1 = 4.
Step 4: Solve the resulting linear equation for x:
2x−1 = 4
2x= 5
x=5
2
Step 5: Therefore, the solution to the exponential equation 32x−1= 81 is
x=5
2.
Question 27
Question
Samantha invests $10,000 in an account that pays 3.5% interest compounded
continuously. How long will it take for her investment to double in value?
18
Solution
Step 1: First, we need to find the formula for continuously compounded interest.
The formula is given by:
A=P·ert
where: - Ais the amount of money accumulated after tyears, - Pis the princi-
pal amount (initial investment), - ris the annual interest rate (expressed as a
decimal), - tis the time the money is invested for, - eis the base of the natural
logarithm, approximately equal to 2.71828.
Step 2: Since Samantha wants her investment to double, we set A= 2Pin
the formula:
2P=P·e0.035t
Step 3: Divide both sides by Pto simplify the equation:
2 = e0.035t
Step 4: Take the natural logarithm of both sides to solve for t:
ln 2 = ln e0.035t
ln 2 = 0.035t·ln e
ln 2 = 0.035t
Step 5: Solve for tby dividing both sides by 0.035:
t=ln 2
0.035
t≈0.6931
0.035
t≈19.802 years
Therefore, it will take approximately 19.802 years for Samantha’s investment
to double in value when compounded continuously at a rate of 3.5%.
Question 28
Question
A population of bacteria doubles every hour. If there are initially 100 bacteria,
how many bacteria will there be after 6 hours?
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Solution
Step 1: Let’s denote the initial number of bacteria as P0= 100 and the growth
factor per hour as r= 2, since the population doubles every hour. The popula-
tion after thours can be represented by the exponential function P(t) = P0·rt.
Step 2: Substitute the given values into the formula to find the population
after 6 hours:
P(6) = 100 ·26
Step 3: Calculate the population after 6 hours:
P(6) = 100 ·26= 100 ·64 = 6400
Step 4: Therefore, after 6 hours, there will be 6400 bacteria in the popula-
tion.
Question 29
Question
A population of bacteria doubles every 3 hours. If there were initially 1000
bacteria in the population, how many bacteria will there be after 12 hours?
Round your answer to the nearest whole number.
Solution
Step 1: Let P(t)represent the population of bacteria after thours. Since the
population doubles every 3 hours, we can express this growth as an exponential
function:
P(t) = 1000 ·2t/3
Step 2: To find the population after 12 hours, substitute t= 12 into the
function:
P(12) = 1000 ·212/3= 1000 ·24= 1000 ·16 = 16000
Step 3: Therefore, after 12 hours, there will be approximately 16,000 bacteria
in the population.
Question 30
Question
Let f(x) = 32xand g(x) = log3(x). Find the value of f(g(9)).
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Solution
Step 1: First, we need to find g(9).
g(9) = log3(9)
= 2 (Since 32= 9)
Step 2: Now, substitute g(9) = 2 into the function f(x)to find f(g(9)).
f(g(9)) = f(2)
= 32(2)
= 34
= 81
Therefore, f(g(9)) = 81.
Question 31
Question
Suppose a certain investment grows according to the formula A(t) = P(1 + r
n)nt,
where A(t)represents the amount of money in the account after tyears, Pis
the principal investment, ris the annual interest rate (expressed as a decimal),
and nis the number of times the interest is compounded per year. If P=$500,
r= 0.05, and n= 4, determine how long it will take for the investment to reach
$800.
Solution
Step 1: Substitute the given values into the formula A(t) = P(1 + r
n)nt to find
the equation that represents the amount of money in the account after tyears:
A(t) = 500 (1 + 0.05
4)4t
Step 2: Set up the equation 800 = 500 (1 + 0.05
4)4t, since we want to find
how long it will take for the investment to reach $800.
Step 3: Divide both sides of the equation by 500 to isolate the exponential
term:
800
500 =(1 + 0.05
4)4t
Step 4: Simplify the left side of the equation:
1.6 = (1 + 0.05
4)4t
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Step 5: Rewrite the equation as an exponential equation:
1.6 = (1 + 0.0125)4t
Step 6: Simplify inside the parentheses:
1.6 = (1.0125)4t
Step 7: Take the natural logarithm of both sides to solve for t:
ln(1.6) = ln((1.0125)4t)
Step 8: Use the property of logarithms that allows us to bring down the
exponent:
ln(1.6) = 4tln(1.0125)
Step 9: Solve for tby dividing both sides by 4 ln(1.0125):
t=ln(1.6)
4 ln(1.0125)
Step 10: Use a calculator to approximate t:
t≈ln(1.6)
4 ln(1.0125) ≈0.4700
4(0.0124) ≈0.4700
0.0496 ≈9.48
Therefore, it will take approximately 9.48 years for the investment to reach
$800.
Question 32
Question
Solve for x:
52x−1=1
125
Solution
Step 1: Rewrite the fraction on the right side as a power of 5:
1
125 = 5−3
Step 2: Substitute 5−3on the right side of the equation:
52x−1= 5−3
Step 3: Since the bases are the same, we can equate the exponents:
2x−1 = −3
Step 4: Solve for x:
2x=−2
x=−1
Step 5: Therefore, the solution to the equation 52x−1=1
125 is x=−1.
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Question 33
Question
Suppose a colony of bacteria triples in population every 6 hours. If the initial
population is 100 bacteria, what will be the population after 24 hours?
Solution
Step 1: Since the population triples every 6 hours, we can express the population
at any time tas P= 100 ·3t/6, where tis the time in hours.
Step 2: To find the population after 24 hours, we substitute t= 24 into the
equation:
P= 100 ·324/6= 100 ·34
Step 3: Calculating 34, we get:
34= 81
Step 4: Finally, we find the population after 24 hours:
P= 100 ·81 = 8100
Therefore, the population after 24 hours will be 8100 bacteria.
Question 34
Question
Solve the exponential equation 3x−2−2(3x) = 1.
Solution
Step 1: Let’s rewrite the equation by factoring out a 3xterm:
3x−2−2(3x) = 1
32·3−2−2(3x) = 1
3−2(32)−2(3x) = 1
1−2(3x) = 1
Step 2: Subtract 1 from both sides to get:
−2(3x) = 0
Step 3: Divide by −2to solve for 3x:
3x= 0
Step 4: Since 3xcannot be equal to 0 for any real value of x, this equation
has no solution.
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Question 35
Question
Solve for x:32x−1= 27
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the equation equal to 33and solve for x.
32x−1= 33
Step 3: Use the properties of exponents to set the exponents equal to each
other.
2x−1 = 3
Step 4: Add 1to both sides of the equation.
2x= 4
Step 5: Divide both sides by 2to solve for x.
x= 2
Therefore, the solution to the equation 32x−1= 27 is x= 2.
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