MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 9
Liberty University
Question 1
Question
Solve the exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,the equation becomes 32x−1= 33.
Step 2: Set the exponents equal to each other.
This gives us the equation 2x−1 = 3.
Step 3: Solve for x.
Adding 1to both sides, we get 2x= 4.
Step 4: Divide by 2 to isolate x.
Dividing by 2yields x= 2.
Step 5: Check that the solution is valid.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
Step 6: Verify the solution.
This simplifies to 33= 33,which is true. Therefore, x= 2 is the solution to the equation.
Question 2
Question
The population of a city is modeled by the function P(t) = 12000 ·e0.02t, where
trepresents the number of years since the population was recorded. Find the
population of the city after 10 years, and determine the rate at which the pop-
ulation is growing at that time.
Solution
Step 1: To find the population of the city after 10 years, we substitute t= 10
into the population function P(t).
P(10) = 12000 ·e0.02·10
Step 2: Simplify the expression by calculating the exponent.
P(10) = 12000 ·e0.2
Step 3: Calculate the value of e0.2.
e0.2≈1.2214
Step 4: Multiply to find the population after 10 years.
P(10) ≈12000 ·1.2214 ≈14656.8
Therefore, the population of the city after 10 years is approximately 14,656.8.
Step 5: To determine the rate at which the population is growing after
10 years, we need to find the derivative of the population function P(t)with
respect to t, which gives us the instantaneous rate of change of the population
with respect to time. dP
dt = 240e0.02t
Step 6: Substitute t= 10 into dP
dt to find the rate of population growth after
10 years.
dP
dt
t=10
= 240e0.02·10
Step 7: Simplify and calculate the rate.
dP
dt
t=10
= 240e0.2≈293.1
Therefore, the rate at which the population is growing after 10 years is
approximately 293.1 individuals per year.
2
Question 3
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 original equation and solve for x.
32x+1 = 32
2x+ 1 = 2
Step 3: Solve for x.
2x= 1
x=1
2
Therefore, the solution to the exponential equation 32x+1 = 9 is x=1
2.
Question 4
Question
Solve for xin the equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the equation 3x−1= 27.
3x−1= 33
Step 3: Since the bases are the same, equate the exponents.
x−1 = 3
Step 4: Solve for xby adding 1to both sides of the equation.
x= 3 + 1
Step 5: Simplify the expression to find the value of x.
x= 4
Therefore, the solution to the equation 3x−1= 27 is x= 4.
3
Question 5
Question
Solve the following exponential equation for x:32x+1 = 5x−3.
Solution
Step 1: Take the natural logarithm of both sides of the equation to eliminate
the exponents:
ln32x+1= ln5x−3
Step 2: Apply the power rule of logarithms to simplify the left side of the
equation:
ln32x+1= (2x+ 1) ln(3)
Step 3: Similarly, apply the power rule of logarithms to simplify the right
side of the equation:
ln5x−3= (x−3) ln(5)
Step 4: Substitute back into the original equation to get:
(2x+ 1) ln(3) = (x−3) ln(5)
Step 5: Expand both sides of the equation:
2xln(3) + ln(3) = xln(5) −3 ln(5)
Step 6: Rearrange the equation to get all terms involving xon one side:
2xln(3) −xln(5) = −3 ln(5) −ln(3)
Step 7: Factor out the x:
x(2 ln(3) −ln(5)) = −3 ln(5) −ln(3)
Step 8: Solve for xby dividing both sides by (2 ln(3) −ln(5)):
x=−3 ln(5) −ln(3)
2 ln(3) −ln(5)
Therefore, the solution to the equation 32x+1 = 5x−3is x=−3 ln(5)−ln(3)
2 ln(3)−ln(5) .
Question 6
Question
Solve the exponential equation 4x−1= 32.
4
Solution
Step 1: Rewrite 32 as a power of 4.
32 = 42
Step 2: Substitute 42for 32 in the original equation.
4x−1= 42
Step 3: Set the exponents equal to each other.
x−1 = 2
Step 4: Solve for x.
x= 2 + 1
x= 3
Step 5: Check the solution.
Substitute x= 3 back into the original equation.
43−1= 32
42= 32
16 = 32
Since 16 = 32, the solution x= 3 is extraneous and the equation has no
solution.
Question 7
Question
Solve the exponential equation log2(x+ 3) −log2(x−3) = 3 for x.
Solution
Step 1: We start by using the properties of logarithms to simplify the equation.
Step 2: The difference of two logarithms with the same base is equal to the
logarithm of the division of their arguments. Thus, we have log2x+3
x−3= 3.
Step 3: Since loga(b) = ccan be rewritten as ac=b, we get 23=x+3
x−3. Step 4:
Simplifying 23gives us 8, so the equation becomes 8 = x+3
x−3. Step 5: Next, we
solve for xby multiplying both sides by x−3to get rid of the denominator. This
gives us 8(x−3) = x+3. Step 6: Expanding the left side, we get 8x−24 = x+3.
Step 7: Combining like terms, we have 7x= 27. Step 8: Finally, we solve for x
by dividing both sides by 7, giving us x=27
7.
Question 8
Question
Samantha invests $5000 in an account with an annual interest rate of 4% that
is compounded continuously. How much will be in the account after 10 years?
5
Solution
Step 1: We can use the formula for compound interest with continuous com-
pounding:
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (the initial amount of money). - ris the annual
interest rate (in decimal form). - tis the time the money is invested for (in
years). - e≈2.71828 is Euler’s number.
Step 2: In this case: - P= $5000 -r= 0.04 -t= 10
Step 3: Substituting the values into the formula:
A= 5000 ·e0.04·10
Step 4: Calculating the exponent:
A= 5000 ·e0.4
Step 5: Evaluating e0.4:
A= 5000 ·1.49182469
Step 6: Calculating the final amount:
A≈7459.12
Therefore, after 10 years, there will be approximately $7459.12 in the ac-
count.
Question 9
Question
Solve for xin the equation 2x+ 3 ·2x−1= 40.
Solution
Step 1: We can rewrite 3·2x−1as 3·2·2x−1= 6 ·2x−1.
Step 2: Substituting this back into the original equation gives us 2x+ 6 ·
2x−1= 40.
Step 3: Factor out 2x−1from both terms to simplify the equation:
2x−1(2 + 6) = 40
Step 4: Simplify the equation further to get:
2x−1·8 = 40
6
Step 5: Divide both sides by 8 to solve for 2x−1:
2x−1= 5
Step 6: To solve for x, we can rewrite 5as 22·23:
2x−1= 22·23
Step 7: Using the properties of exponents, we can simplify the equation to:
2x−1= 2x+2
Step 8: Setting the exponents equal to each other, we have:
x−1 = x+ 2
Step 9: Solving the equation for x, we get:
−1 = 2
Step 10: Since −1= 2, there is no solution to the equation 2x+3·2x−1= 40.
Question 10
Question
Solve the following exponential equation for x:
52x+1 = 25
Solution
Step 1: Rewrite both sides of the equation with the same base.
52x+1 = 25
52x+1 = 52
Step 2: Set the exponents equal to each other.
2x+ 1 = 2
Step 3: Solve for x.
2x+ 1 = 2
2x= 1
x=1
2
7
Step 4: Check the solution by substituting x=1
2back into the original
equation.
52( 1
2)+1 = 25
52+1 = 25
53= 25
125 = 25
Since the final step leads to a false statement, there is no solution to the
equation.
Question 11
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the given equation.
32x−1= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x−1 = 3
Step 4: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 5: Check the solution by substituting x= 2 back into the original
equation.
32(2)−1= 27
34−1= 33
33= 27
Therefore, the solution to the exponential equation 32x−1= 27 is x= 2.
8
Question 12
Question
The population of a certain city is modeled by the function P(t) = 10,000·e0.02t,
where tis the number of years since the start of 2020. Find and interpret the
population in the year 2030.
Solution
Step 1: To find the population in the year 2030, we need to substitute t=
2030 −2020 = 10 into the population function P(t).
P(10) = 10,000 ·e0.02·10
Step 2: Calculate the value of P(10).
P(10) = 10,000 ·e0.2
P(10) = 10,000 ·e0.2
P(10) ≈12214.29
Step 3: Interpretation: The population of the city in the year 2030 is ap-
proximately 12,214.29 people.
Question 13
Question
Solve for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Step 2: Solve for xusing the properties of
exponents and logarithms.
Step 1: Rewrite 8as a power of 2.
8 = 23
Step 2: Solve for xusing the properties of exponents and logarithms.
23x−1= 23
Since the bases are the same, we can equate the exponents:
3x−1 = 3
Solving for x:
3x= 4
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
9
Question 14
Question
You have just won a lottery that promises to pay you 80,000ayearf orthenext10years.However, youhavetheoptiontotakealumpsumpaymentnowinstead.T helumpsumpaymentisdeterminedusinganannualinterestrateof 5%.W hatistheminimumlumpsumamountyoushouldacceptnowif youcaninvestthemoneyat5%interestcompoundedannually?
Solution
Step 1: Calculate the present value of the annuity using the formula for the
present value of an annuity:
P V =P×1−1
(1 + r)n/r
Where, - P V is the present value of the annuity (the lump sum payment), -
Pis the annual payment amount (80,000),−ris the annual interest rate (5 or
0.05), - nis the number of years the annuity will be paid (10 years).
Plugging in the values, we get:
P V = 80000 ×1−1
(1 + 0.05)10 /0.05
Step 2: Calculate the present value of the annuity:
P V = 80000 ×1−1
1.0510 /0.05
P V = 80000 ×1−1
1.62889462677/0.05
P V = 80000 ×(1 −0.613913253) /0.05
P V = 80000 ×0.386086747/0.05
P V = 30887.39976
Therefore, the present value of the annuity, or the minimum lump sum
amount you should accept now, is $30,887.40.
Question 15
Question
Suppose the population of a city is modeled by the function P(t) = 5000 ·1.03t,
where trepresents the number of years after 2020. Find the year when the
population is expected to reach 8000.
10
Solution
Step 1: Set up the equation based on the information given. We want to find
the year when the population reaches 8000, so we set P(t) = 8000:
5000 ·1.03t= 8000
Step 2: Solve for tby dividing both sides by 5000:
1.03t=8000
5000
1.03t= 1.6
Step 3: Take the natural logarithm of both sides to solve for t:
ln1.03t= ln(1.6)
tln(1.03) = ln(1.6)
Step 4: Solve for tby dividing both sides by ln(1.03):
t=ln(1.6)
ln(1.03)
Step 5: Use a calculator to find the approximate value of t:
t≈ln(1.6)
ln(1.03) ≈12.52
Step 6: Add tto the base year 2020 to find the year when the population is
expected to reach 8000:
2020 + 12.52 ≈2032.52
Therefore, the population is expected to reach 8000 around the year 2033.
Question 16
Question
Solve the exponential equation 3x−2−3x−1= 8.
Solution
Step 1: Let’s rewrite the equation using the properties of exponents. We know
that am·an=am+n. So we can rewrite the given equation as 3x−2·3−1= 8.
Step 2: Simplifying the right side, we get 3x−1= 8.
Step 3: Now we can rewrite 8 as a power of 3. Since 32= 9, we have
8 = 3log38= 3 log 8
log 3 = 3 3
log 3 .
11
Step 4: Substituting the value of 8 back into our equation, we get 3x−1=
33
log 3 .
Step 5: Since the bases are the same, we can set the exponents equal to each
other. Thus, x−1 = 3
log 3 .
Step 6: To solve for x, we add 1 to both sides of the equation, giving us
x= 1 + 3
log 3 . Hence, the solution to the exponential equation is x= 1 + 3
log 3 .
Question 17
Question
Solve the exponential equation 2x+3 −2x+2 = 24.
Solution
Step 1: Rewrite the equation using the properties of exponents. Step 2: Apply
the properties of exponents to simplify the equation. Step 3: Solve for xby
isolating 2x. Step 4: Check the solution to ensure it is valid.
Step 1: Rewrite the equation using the properties of exponents.
We rewrite 2x+3 −2x+2 = 24 as 2·2x−2·2x= 24.
Step 2: Apply the properties of exponents to simplify the equation.
Simplifying further, we get 2x−2x= 24.
Step 3: Solve for xby isolating 2x.
Since 2x−2x= 0, we have 0 = 24 which is not possible. Therefore, there is
no solution to the equation 2x+3 −2x+2 = 24.
Question 18
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 exponents equal to each other.
2x+ 1 = 3
Step 3: Solve for x.
2x+ 1 = 3
2x= 2
x= 1
12
Step 4: Check the solution.
32(1)+1 = 33= 27
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 19
Question
Solve the following exponential equation for x:23x= 8.
Solution
Step 1: Rewrite 8 as a power of 2. Step 2: Solve for xby equating the exponents.
Question 20
Question
Solve for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3. We know that 33= 27, so 27 = 33.
Step 2: Substitute 27 with 33in the equation 32x−1= 27. We get 32x−1= 33.
Step 3: Since the bases are equal, we can set the exponents equal to each
other. So, 2x−1 = 3.
Step 4: Add 1 to both sides of the equation to isolate the term with x:
2x= 4.
Step 5: Divide by 2 on both sides to solve for x:x= 2.
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 21
Question
Solve the logarithmic equation: log5(3x+ 1) + log5(x−2) = 2.
Solution
Step 1: Apply the product rule of logarithms to combine the logarithms on the
left side.
log5[(3x+ 1)(x−2)] = 2
13
Step 2: Rewrite the equation in exponential form. Remember that loga(b) =
cis equivalent to ac=b.
52= (3x+ 1)(x−2)
Step 3: Simplify the equation by squaring 5.
25 = 3x2−6x+x−2
Step 4: Combine like terms and set the equation equal to zero.
3x2−5x−27 = 0
Step 5: Factor the quadratic equation.
(3x+ 9)(x−3) = 0
Step 6: Solve for x by setting each factor to zero.
3x+ 9 = 0 or x−3 = 0
Step 7: Solve for x in each equation to find the possible values.
x=−3or x= 3
Step 8: Check for extraneous solutions by substituting the values back into
the original logarithmic equation.
x=−3 : log5(3(−3) + 1) + log5((−3) −2) = log5(−8) + undefined
x= 3 : log5(3(3) + 1) + log5(3 −2) = log5(10) + log5(1) = 1 + 0 = 2
Step 9: Therefore, the solution to the logarithmic equation is x= 3.
Question 22
Question
Suppose a bacteria culture starts with 100 bacteria and grows at a rate of 8%
per hour. Write an exponential growth model for the number of bacteria after
thours and determine how many bacteria will be present after 5 hours.
Solution
Step 1: Let N(t)represent the number of bacteria after thours. Since the pop-
ulation is growing at a rate of 8% per hour, the growth model can be expressed
as:
N(t) = 100 ×(1 + 0.08)t
Step 2: Simplify the model to get a final formula for N(t):
N(t) = 100 ×(1.08)t
14
Step 3: To find the number of bacteria after 5 hours, substitute t= 5 into
the formula:
N(5) = 100 ×(1.08)5
N(5) = 100 ×1.46933
N(5) ≈146.93
Therefore, there will be approximately 146.93 bacteria present after 5 hours.
Question 23
Question
Suppose the population of a town is modeled by the function P(t) = 5000·e0.02t,
where trepresents the number of years since the initial population count was
taken. Find the population after 10 years to the nearest whole number.
Solution
Step 1: Substitute t= 10 into the given function to find the population after
10 years.
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression inside the exponent.
P(10) = 5000 ·e0.2
Step 3: Calculate the exponential term.
P(10) = 5000 ·1.221402
Step 4: Multiply to find the population after 10 years.
P(10) ≈6107
Therefore, the population of the town after 10 years is approximately 6107
people.
Question 24
Question
Solve the exponential equation 32x+1 −2·3x= 5 for x.
15
Solution
Step 1: Let’s rewrite the given equation into a more manageable form by using
a substitution. Let u= 3x. Then, the equation becomes 3u2−2u= 5.
Step 2: Rearrange the equation 3u2−2u= 5 into a quadratic form by
moving all terms to one side of the equation. We get 3u2−2u−5 = 0.
Step 3: Now, we can solve the quadratic equation 3u2−2u−5=0for u
using the quadratic formula: u=−(−2)±√(−2)2−4·3·(−5)
2·3.
Step 4: Simplify the quadratic formula to find the solutions for u:u=
2±√4+60
6. Thus, u=2±√64
6.
Step 5: Continuing to simplify, we get two solutions: u=2±8
6. So, u=10
6
or u=−6
6.
Step 6: Therefore, u=5
3or u=−1. Recall that u= 3x, so we have 3x=5
3
or 3x=−1. However, 3xcannot be negative, so we focus on the equation
3x=5
3.
Step 7: To solve 3x=5
3, take the natural logarithm of both sides: ln(3x) =
ln 5
3.
Step 8: Apply the logarithmic property to bring down the exponent: xln(3) =
ln 5
3.
Step 9: Solve for x:x=ln(5
3)
ln(3) . This is the exact solution for x.
Question 25
Question
Suppose a bacteria population doubles every 6 hours. If there are initially 100
bacteria, how many bacteria will there be after 24 hours?
Solution
Step 1: Find the growth factor from the doubling time.
Given that the population doubles every 6 hours, we can find the growth
factor, r, using the formula:
r= 2 1
doubling time = 2 1
6
Step 2: Calculate the growth factor.
r= 21
6= 21
6= 21
6= 20.1667 ≈1.1225
Step 3: Use the growth factor to find the population after 24 hours.
With an initial population of 100 bacteria and a growth factor of 1.1225, the
population after 24 hours can be calculated using the formula for exponential
growth:
Population after 24 hours =Initial population ×(1.1225) time (in hours)
doubling time
16
Population after 24 hours = 100 ×(1.1225)24
6
Population after 24 hours = 100 ×(1.1225)4
Population after 24 hours = 100 ×(1.601)4= 100 ×2.564 ≈256.4
Therefore, after 24 hours, there will be approximately 256 bacteria in the
population.
Question 26
Question
The population of a city is modeled by the function P(t) = 50,000·1.02t, where
trepresents the number of years since 2020. Calculate the population of the
city in the year 2040.
Solution
Step 1: Determine the number of years from 2020 to 2040. We need to find t
when the year is 2040.
2040 −2020 = 20 years
Step 2: Substitute the value of tinto the population function.
P(20) = 50,000 ·1.0220
Step 3: Calculate 1.0220.
1.0220 ≈1.485
Step 4: Substitute this value back into the population function.
P(20) = 50,000 ·1.485
P(20) = 74250
So, the population of the city in the year 2040 will be 74,250.
Question 27
Question
The population of a city is modeled by the exponential function P(t) = 5000 ·
1.02t, where trepresents the number of years since 2020. What will the popu-
lation be in the year 2030? Round your answer to the nearest whole number.
17
Solution
Step 1: Substitute t= 10 into the function to find the population in the year
2030.
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210.
1.0210 ≈1.218994
Step 3: Substitute the value of 1.0210 back into the equation.
P(10) = 5000 ·1.218994
Step 4: Calculate 5000 ·1.218994.
5000 ·1.218994 ≈6094
Therefore, the population of the city in the year 2030 will be approximately
6094.
Question 28
Question
A certain radioactive substance decays according to the formula A=A0e−kt,
where Ais the amount of substance remaining after tyears, A0is the initial
amount of substance, and kis a constant. If 80% of the substance decays in 30
years, what is the half-life of the substance?
Solution
Step 1: To find the half-life of the substance, we first need to determine the
value of kin the exponential decay formula A=A0e−kt .
Let A0be the initial amount of substance and Abe the amount remaining
after tyears. We are given that 80% of the substance decays in 30 years, so this
means that after 30 years, only 20% of the substance is left.
Step 2: Since only 20% of the substance remains after 30 years, we have:
0.20A0=A0e−30k
Step 3: Simplifying the equation, we find:
e−30k= 0.20
Step 4: Taking the natural logarithm of both sides gives:
lne−30k= ln(0.20)
18
Step 5: Using the property of logarithms that ln(ex) = x, we get:
−30k= ln(0.20)
Step 6: Solving for k, we find:
k=ln(0.20)
−30 ≈0.0231
Step 7: Now that we have the value of k, we can find the half-life of the
substance. The half-life is the time required for half of the substance to decay.
Step 8: Let A=1
2A0(half of the initial amount). We can plug this into the
exponential decay formula to find the half-life:
1
2A0=A0e−0.0231t
Step 9: Dividing by A0and solving for t, we get:
e0.0231t= 2
Step 10: Taking the natural logarithm of both sides gives:
lne0.0231t= ln(2)
Step 11: Simplifying, we find:
0.0231t= ln(2)
Step 12: Solving for t, we get:
t=ln(2)
0.0231 ≈30.03 years
Therefore, the half-life of the substance is approximately 30.03 years.
Question 29
Question
Given the exponential function P(t) = 200e0.03t, where trepresents time in
years, determine the time it takes for an initial investment of 200togrowto220.
Round your answer to the nearest hundredth.
Solution
Step 1: Set up the equation P(t) = 220 and solve for t.
200e0.03t= 220
e0.03t=220
200
e0.03t= 1.1
19
Step 2: Take the natural logarithm of both sides to solve for t.
lne0.03t= ln(1.1)
0.03t= ln(1.1)
t=ln(1.1)
0.03
Step 3: Use a calculator to find the approximate value of trounded to the
nearest hundredth.
t≈ln(1.1)
0.03
t≈0.09531
0.03
t≈3.18
Therefore, it takes approximately 3.18 years for the initial investment of
200togrowto220.
Question 30
Question
Solve the exponential equation 32x−1−5·3x+ 6 = 0 for x.
Solution
Step 1: Let’s rewrite the exponential equation as a quadratic equation:
(3x)2−5·3x+ 6 = 0
Step 2: Let y= 3x. Then the quadratic equation becomes:
y2−5y+ 6 = 0
Step 3: Now, let’s factor the quadratic equation:
(y−2)(y−3) = 0
Step 4: Set each factor to zero and solve for y:
y−2 = 0 =⇒y= 2 or y−3 = 0 =⇒y= 3
Step 5: Now that we have the possible values for y, let’s substitute back to
solve for x: For y= 2:
3x= 2
⇒x= log3(2)
For y= 3:
3x= 3
⇒x= 1
Step 6: Therefore, the solutions to the exponential equation are x= log3(2)
and x= 1.
20
Question 31
Question
Suppose the population of a city grows exponentially at a rate of 3.5% per year.
If the current population is 100,000, what will the population be in 10 years?
Solution
Step 1: To find the population in 10 years, we can use the formula for exponential
growth:
P(t) = P0×(1 + r)t
where: - P(t)is the population after time t, - P0is the initial population, - ris
the growth rate, and - tis the time period.
Step 2: Substituting the given values into the formula, we have:
P(10) = 100,000 ×(1 + 0.035)10
Step 3: Calculating the population after 10 years:
P(10) = 100,000 ×(1.035)10
P(10) = 100,000 ×1.418519
P(10) = 141,851.9
Step 4: Therefore, the population of the city will be approximately 141,852
in 10 years.
Question 32
Question
A company has determined that its monthly profit can be modeled by the func-
tion P(x) = 5000(1.10)x−5000, where xrepresents the number of months since
the company began operating. Find the number of months it will take for the
company to reach a monthly profit of 30,000.
Solution
Step 1: Set up the equation using the given information. We want to find the
number of months xwhen the monthly profit P(x)is 30,000.T hus, wehave :
30,000 = 5000(1.10)x−5000
Step 2: Solve for xby isolating the exponential term. We first add 5000 to
both sides of the equation:
35,000 = 5000(1.10)x
21
Step 3: Divide both sides by 5000 to isolate the exponential term:
7 = 1.10x
Step 4: Take the logarithm of both sides to solve for x. Let’s take the log1.10
of both sides:
log1.10 7 = log1.10 1.10x
Step 5: By the logarithm properties, the exponent can be moved down as a
coefficient:
log1.10 7 = x
Step 6: Finally, calculate the value of x. Using a calculator, we find:
x≈5.94
Therefore, it will take approximately 5.94 months for the company to reach
a monthly profit of 30,000.
Question 33
Question
Let f(x)=3·2xand g(x) = log2(x). Find the values of xthat satisfy f(x)>
g(x).
Solution
Step 1: We start by setting up the inequality f(x)> g(x)and using the given
functions f(x)and g(x):
3·2x>log2(x)
Step 2: To simplify the inequality, we can rewrite log2(x)as 2log2(x)=x:
3·2x> x
Step 3: Next, we can rewrite 3as 3·20and then simplify the inequality
further:
3·2x−0> x
2x+1 > x
Step 4: To solve this inequality, we can set up a graph for y= 2x+1 and
y=xon the same set of axes:
Step 5: By analyzing the graph, we see that the two functions intersect at
x= 1.
Step 6: To the determine the values of xfor which f(x)> g(x), we need to
consider the regions where f(x)is greater than g(x):
In the interval (−∞,1),2x+1 > x is true.
In the interval (1,∞),2x+1 < x is true.
Step 7: Therefore, the values of xthat satisfy f(x)> g(x)are x∈(−∞,1).
22
Question 34
Question
A certain radioactive substance decays according to the formula Q(t) = Q0·
e−0.025t, where Q(t)represents the amount of the substance remaining after t
years and Q0is the initial amount present. If 500 grams of the substance are
initially present, how many grams will be left after 10 years?
Solution
Step 1: Given that the initial amount Q0= 500 grams and the formula for the
decay is Q(t) = Q0·e−0.025t, we can substitute these values into the formula:
Q(t) = 500 ·e−0.025t
Step 2: To find the amount of the substance remaining after 10 years, we
substitute t= 10 into the formula:
Q(10) = 500 ·e−0.025·10
Step 3: Calculate the value of Q(10):
Q(10) = 500 ·e−0.25
Step 4: Evaluate e−0.25 using a calculator:
Q(10) = 500 ·0.77880078307
Step 5: Multiply to find the amount remaining after 10 years:
Q(10) ≈389.400391535
Therefore, after 10 years, there will be approximately 389.4 grams of the
substance remaining.
Question 35
Question
Solve the exponential equation 3x= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
23
Question 2
Question
The population of a city is modeled by the function P(t) = 12000 ·e0.02t, where
trepresents the number of years since the population was recorded. Find the
population of the city after 10 years, and determine the rate at which the pop-
ulation is growing at that time.
Solution
Step 1: To find the population of the city after 10 years, we substitute t= 10
into the population function P(t).
P(10) = 12000 ·e0.02·10
Step 2: Simplify the expression by calculating the exponent.
P(10) = 12000 ·e0.2
Step 3: Calculate the value of e0.2.
e0.2≈1.2214
Step 4: Multiply to find the population after 10 years.
P(10) ≈12000 ·1.2214 ≈14656.8
Therefore, the population of the city after 10 years is approximately 14,656.8.
Step 5: To determine the rate at which the population is growing after
10 years, we need to find the derivative of the population function P(t)with
respect to t, which gives us the instantaneous rate of change of the population
with respect to time. dP
dt = 240e0.02t
Step 6: Substitute t= 10 into dP
dt to find the rate of population growth after
10 years.
dP
dt
t=10
= 240e0.02·10
Step 7: Simplify and calculate the rate.
dP
dt
t=10
= 240e0.2≈293.1
Therefore, the rate at which the population is growing after 10 years is
approximately 293.1 individuals per year.
2
Question 3
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 original equation and solve for x.
32x+1 = 32
2x+ 1 = 2
Step 3: Solve for x.
2x= 1
x=1
2
Therefore, the solution to the exponential equation 32x+1 = 9 is x=1
2.
Question 4
Question
Solve for xin the equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the equation 3x−1= 27.
3x−1= 33
Step 3: Since the bases are the same, equate the exponents.
x−1 = 3
Step 4: Solve for xby adding 1to both sides of the equation.
x= 3 + 1
Step 5: Simplify the expression to find the value of x.
x= 4
Therefore, the solution to the equation 3x−1= 27 is x= 4.
3
Question 5
Question
Solve the following exponential equation for x:32x+1 = 5x−3.
Solution
Step 1: Take the natural logarithm of both sides of the equation to eliminate
the exponents:
ln32x+1= ln5x−3
Step 2: Apply the power rule of logarithms to simplify the left side of the
equation:
ln32x+1= (2x+ 1) ln(3)
Step 3: Similarly, apply the power rule of logarithms to simplify the right
side of the equation:
ln5x−3= (x−3) ln(5)
Step 4: Substitute back into the original equation to get:
(2x+ 1) ln(3) = (x−3) ln(5)
Step 5: Expand both sides of the equation:
2xln(3) + ln(3) = xln(5) −3 ln(5)
Step 6: Rearrange the equation to get all terms involving xon one side:
2xln(3) −xln(5) = −3 ln(5) −ln(3)
Step 7: Factor out the x:
x(2 ln(3) −ln(5)) = −3 ln(5) −ln(3)
Step 8: Solve for xby dividing both sides by (2 ln(3) −ln(5)):
x=−3 ln(5) −ln(3)
2 ln(3) −ln(5)
Therefore, the solution to the equation 32x+1 = 5x−3is x=−3 ln(5)−ln(3)
2 ln(3)−ln(5) .
Question 6
Question
Solve the exponential equation 4x−1= 32.
4
Solution
Step 1: Rewrite 32 as a power of 4.
32 = 42
Step 2: Substitute 42for 32 in the original equation.
4x−1= 42
Step 3: Set the exponents equal to each other.
x−1 = 2
Step 4: Solve for x.
x= 2 + 1
x= 3
Step 5: Check the solution.
Substitute x= 3 back into the original equation.
43−1= 32
42= 32
16 = 32
Since 16 = 32, the solution x= 3 is extraneous and the equation has no
solution.
Question 7
Question
Solve the exponential equation log2(x+ 3) −log2(x−3) = 3 for x.
Solution
Step 1: We start by using the properties of logarithms to simplify the equation.
Step 2: The difference of two logarithms with the same base is equal to the
logarithm of the division of their arguments. Thus, we have log2x+3
x−3= 3.
Step 3: Since loga(b) = ccan be rewritten as ac=b, we get 23=x+3
x−3. Step 4:
Simplifying 23gives us 8, so the equation becomes 8 = x+3
x−3. Step 5: Next, we
solve for xby multiplying both sides by x−3to get rid of the denominator. This
gives us 8(x−3) = x+3. Step 6: Expanding the left side, we get 8x−24 = x+3.
Step 7: Combining like terms, we have 7x= 27. Step 8: Finally, we solve for x
by dividing both sides by 7, giving us x=27
7.
Question 8
Question
Samantha invests $5000 in an account with an annual interest rate of 4% that
is compounded continuously. How much will be in the account after 10 years?
5
Solution
Step 1: We can use the formula for compound interest with continuous com-
pounding:
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (the initial amount of money). - ris the annual
interest rate (in decimal form). - tis the time the money is invested for (in
years). - e≈2.71828 is Euler’s number.
Step 2: In this case: - P= $5000 -r= 0.04 -t= 10
Step 3: Substituting the values into the formula:
A= 5000 ·e0.04·10
Step 4: Calculating the exponent:
A= 5000 ·e0.4
Step 5: Evaluating e0.4:
A= 5000 ·1.49182469
Step 6: Calculating the final amount:
A≈7459.12
Therefore, after 10 years, there will be approximately $7459.12 in the ac-
count.
Question 9
Question
Solve for xin the equation 2x+ 3 ·2x−1= 40.
Solution
Step 1: We can rewrite 3·2x−1as 3·2·2x−1= 6 ·2x−1.
Step 2: Substituting this back into the original equation gives us 2x+ 6 ·
2x−1= 40.
Step 3: Factor out 2x−1from both terms to simplify the equation:
2x−1(2 + 6) = 40
Step 4: Simplify the equation further to get:
2x−1·8 = 40
6
Step 5: Divide both sides by 8 to solve for 2x−1:
2x−1= 5
Step 6: To solve for x, we can rewrite 5as 22·23:
2x−1= 22·23
Step 7: Using the properties of exponents, we can simplify the equation to:
2x−1= 2x+2
Step 8: Setting the exponents equal to each other, we have:
x−1 = x+ 2
Step 9: Solving the equation for x, we get:
−1 = 2
Step 10: Since −1= 2, there is no solution to the equation 2x+3·2x−1= 40.
Question 10
Question
Solve the following exponential equation for x:
52x+1 = 25
Solution
Step 1: Rewrite both sides of the equation with the same base.
52x+1 = 25
52x+1 = 52
Step 2: Set the exponents equal to each other.
2x+ 1 = 2
Step 3: Solve for x.
2x+ 1 = 2
2x= 1
x=1
2
7
Step 4: Check the solution by substituting x=1
2back into the original
equation.
52( 1
2)+1 = 25
52+1 = 25
53= 25
125 = 25
Since the final step leads to a false statement, there is no solution to the
equation.
Question 11
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the given equation.
32x−1= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x−1 = 3
Step 4: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 5: Check the solution by substituting x= 2 back into the original
equation.
32(2)−1= 27
34−1= 33
33= 27
Therefore, the solution to the exponential equation 32x−1= 27 is x= 2.
8
Question 12
Question
The population of a certain city is modeled by the function P(t) = 10,000·e0.02t,
where tis the number of years since the start of 2020. Find and interpret the
population in the year 2030.
Solution
Step 1: To find the population in the year 2030, we need to substitute t=
2030 −2020 = 10 into the population function P(t).
P(10) = 10,000 ·e0.02·10
Step 2: Calculate the value of P(10).
P(10) = 10,000 ·e0.2
P(10) = 10,000 ·e0.2
P(10) ≈12214.29
Step 3: Interpretation: The population of the city in the year 2030 is ap-
proximately 12,214.29 people.
Question 13
Question
Solve for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Step 2: Solve for xusing the properties of
exponents and logarithms.
Step 1: Rewrite 8as a power of 2.
8 = 23
Step 2: Solve for xusing the properties of exponents and logarithms.
23x−1= 23
Since the bases are the same, we can equate the exponents:
3x−1 = 3
Solving for x:
3x= 4
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
9
Question 14
Question
You have just won a lottery that promises to pay you 80,000ayearf orthenext10years.However, youhavetheoptiontotakealumpsumpaymentnowinstead.T helumpsumpaymentisdeterminedusinganannualinterestrateof 5%.W hatistheminimumlumpsumamountyoushouldacceptnowif youcaninvestthemoneyat5%interestcompoundedannually?
Solution
Step 1: Calculate the present value of the annuity using the formula for the
present value of an annuity:
P V =P×1−1
(1 + r)n/r
Where, - P V is the present value of the annuity (the lump sum payment), -
Pis the annual payment amount (80,000),−ris the annual interest rate (5 or
0.05), - nis the number of years the annuity will be paid (10 years).
Plugging in the values, we get:
P V = 80000 ×1−1
(1 + 0.05)10 /0.05
Step 2: Calculate the present value of the annuity:
P V = 80000 ×1−1
1.0510 /0.05
P V = 80000 ×1−1
1.62889462677/0.05
P V = 80000 ×(1 −0.613913253) /0.05
P V = 80000 ×0.386086747/0.05
P V = 30887.39976
Therefore, the present value of the annuity, or the minimum lump sum
amount you should accept now, is $30,887.40.
Question 15
Question
Suppose the population of a city is modeled by the function P(t) = 5000 ·1.03t,
where trepresents the number of years after 2020. Find the year when the
population is expected to reach 8000.
10
Solution
Step 1: Set up the equation based on the information given. We want to find
the year when the population reaches 8000, so we set P(t) = 8000:
5000 ·1.03t= 8000
Step 2: Solve for tby dividing both sides by 5000:
1.03t=8000
5000
1.03t= 1.6
Step 3: Take the natural logarithm of both sides to solve for t:
ln1.03t= ln(1.6)
tln(1.03) = ln(1.6)
Step 4: Solve for tby dividing both sides by ln(1.03):
t=ln(1.6)
ln(1.03)
Step 5: Use a calculator to find the approximate value of t:
t≈ln(1.6)
ln(1.03) ≈12.52
Step 6: Add tto the base year 2020 to find the year when the population is
expected to reach 8000:
2020 + 12.52 ≈2032.52
Therefore, the population is expected to reach 8000 around the year 2033.
Question 16
Question
Solve the exponential equation 3x−2−3x−1= 8.
Solution
Step 1: Let’s rewrite the equation using the properties of exponents. We know
that am·an=am+n. So we can rewrite the given equation as 3x−2·3−1= 8.
Step 2: Simplifying the right side, we get 3x−1= 8.
Step 3: Now we can rewrite 8 as a power of 3. Since 32= 9, we have
8 = 3log38= 3 log 8
log 3 = 3 3
log 3 .
11
Step 4: Substituting the value of 8 back into our equation, we get 3x−1=
33
log 3 .
Step 5: Since the bases are the same, we can set the exponents equal to each
other. Thus, x−1 = 3
log 3 .
Step 6: To solve for x, we add 1 to both sides of the equation, giving us
x= 1 + 3
log 3 . Hence, the solution to the exponential equation is x= 1 + 3
log 3 .
Question 17
Question
Solve the exponential equation 2x+3 −2x+2 = 24.
Solution
Step 1: Rewrite the equation using the properties of exponents. Step 2: Apply
the properties of exponents to simplify the equation. Step 3: Solve for xby
isolating 2x. Step 4: Check the solution to ensure it is valid.
Step 1: Rewrite the equation using the properties of exponents.
We rewrite 2x+3 −2x+2 = 24 as 2·2x−2·2x= 24.
Step 2: Apply the properties of exponents to simplify the equation.
Simplifying further, we get 2x−2x= 24.
Step 3: Solve for xby isolating 2x.
Since 2x−2x= 0, we have 0 = 24 which is not possible. Therefore, there is
no solution to the equation 2x+3 −2x+2 = 24.
Question 18
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 exponents equal to each other.
2x+ 1 = 3
Step 3: Solve for x.
2x+ 1 = 3
2x= 2
x= 1
12
Step 4: Check the solution.
32(1)+1 = 33= 27
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 19
Question
Solve the following exponential equation for x:23x= 8.
Solution
Step 1: Rewrite 8 as a power of 2. Step 2: Solve for xby equating the exponents.
Question 20
Question
Solve for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3. We know that 33= 27, so 27 = 33.
Step 2: Substitute 27 with 33in the equation 32x−1= 27. We get 32x−1= 33.
Step 3: Since the bases are equal, we can set the exponents equal to each
other. So, 2x−1 = 3.
Step 4: Add 1 to both sides of the equation to isolate the term with x:
2x= 4.
Step 5: Divide by 2 on both sides to solve for x:x= 2.
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 21
Question
Solve the logarithmic equation: log5(3x+ 1) + log5(x−2) = 2.
Solution
Step 1: Apply the product rule of logarithms to combine the logarithms on the
left side.
log5[(3x+ 1)(x−2)] = 2
13
Step 2: Rewrite the equation in exponential form. Remember that loga(b) =
cis equivalent to ac=b.
52= (3x+ 1)(x−2)
Step 3: Simplify the equation by squaring 5.
25 = 3x2−6x+x−2
Step 4: Combine like terms and set the equation equal to zero.
3x2−5x−27 = 0
Step 5: Factor the quadratic equation.
(3x+ 9)(x−3) = 0
Step 6: Solve for x by setting each factor to zero.
3x+ 9 = 0 or x−3 = 0
Step 7: Solve for x in each equation to find the possible values.
x=−3or x= 3
Step 8: Check for extraneous solutions by substituting the values back into
the original logarithmic equation.
x=−3 : log5(3(−3) + 1) + log5((−3) −2) = log5(−8) + undefined
x= 3 : log5(3(3) + 1) + log5(3 −2) = log5(10) + log5(1) = 1 + 0 = 2
Step 9: Therefore, the solution to the logarithmic equation is x= 3.
Question 22
Question
Suppose a bacteria culture starts with 100 bacteria and grows at a rate of 8%
per hour. Write an exponential growth model for the number of bacteria after
thours and determine how many bacteria will be present after 5 hours.
Solution
Step 1: Let N(t)represent the number of bacteria after thours. Since the pop-
ulation is growing at a rate of 8% per hour, the growth model can be expressed
as:
N(t) = 100 ×(1 + 0.08)t
Step 2: Simplify the model to get a final formula for N(t):
N(t) = 100 ×(1.08)t
14
Step 3: To find the number of bacteria after 5 hours, substitute t= 5 into
the formula:
N(5) = 100 ×(1.08)5
N(5) = 100 ×1.46933
N(5) ≈146.93
Therefore, there will be approximately 146.93 bacteria present after 5 hours.
Question 23
Question
Suppose the population of a town is modeled by the function P(t) = 5000·e0.02t,
where trepresents the number of years since the initial population count was
taken. Find the population after 10 years to the nearest whole number.
Solution
Step 1: Substitute t= 10 into the given function to find the population after
10 years.
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression inside the exponent.
P(10) = 5000 ·e0.2
Step 3: Calculate the exponential term.
P(10) = 5000 ·1.221402
Step 4: Multiply to find the population after 10 years.
P(10) ≈6107
Therefore, the population of the town after 10 years is approximately 6107
people.
Question 24
Question
Solve the exponential equation 32x+1 −2·3x= 5 for x.
15
Solution
Step 1: Let’s rewrite the given equation into a more manageable form by using
a substitution. Let u= 3x. Then, the equation becomes 3u2−2u= 5.
Step 2: Rearrange the equation 3u2−2u= 5 into a quadratic form by
moving all terms to one side of the equation. We get 3u2−2u−5 = 0.
Step 3: Now, we can solve the quadratic equation 3u2−2u−5=0for u
using the quadratic formula: u=−(−2)±√(−2)2−4·3·(−5)
2·3.
Step 4: Simplify the quadratic formula to find the solutions for u:u=
2±√4+60
6. Thus, u=2±√64
6.
Step 5: Continuing to simplify, we get two solutions: u=2±8
6. So, u=10
6
or u=−6
6.
Step 6: Therefore, u=5
3or u=−1. Recall that u= 3x, so we have 3x=5
3
or 3x=−1. However, 3xcannot be negative, so we focus on the equation
3x=5
3.
Step 7: To solve 3x=5
3, take the natural logarithm of both sides: ln(3x) =
ln 5
3.
Step 8: Apply the logarithmic property to bring down the exponent: xln(3) =
ln 5
3.
Step 9: Solve for x:x=ln(5
3)
ln(3) . This is the exact solution for x.
Question 25
Question
Suppose a bacteria population doubles every 6 hours. If there are initially 100
bacteria, how many bacteria will there be after 24 hours?
Solution
Step 1: Find the growth factor from the doubling time.
Given that the population doubles every 6 hours, we can find the growth
factor, r, using the formula:
r= 2 1
doubling time = 2 1
6
Step 2: Calculate the growth factor.
r= 21
6= 21
6= 21
6= 20.1667 ≈1.1225
Step 3: Use the growth factor to find the population after 24 hours.
With an initial population of 100 bacteria and a growth factor of 1.1225, the
population after 24 hours can be calculated using the formula for exponential
growth:
Population after 24 hours =Initial population ×(1.1225) time (in hours)
doubling time
16
Population after 24 hours = 100 ×(1.1225)24
6
Population after 24 hours = 100 ×(1.1225)4
Population after 24 hours = 100 ×(1.601)4= 100 ×2.564 ≈256.4
Therefore, after 24 hours, there will be approximately 256 bacteria in the
population.
Question 26
Question
The population of a city is modeled by the function P(t) = 50,000·1.02t, where
trepresents the number of years since 2020. Calculate the population of the
city in the year 2040.
Solution
Step 1: Determine the number of years from 2020 to 2040. We need to find t
when the year is 2040.
2040 −2020 = 20 years
Step 2: Substitute the value of tinto the population function.
P(20) = 50,000 ·1.0220
Step 3: Calculate 1.0220.
1.0220 ≈1.485
Step 4: Substitute this value back into the population function.
P(20) = 50,000 ·1.485
P(20) = 74250
So, the population of the city in the year 2040 will be 74,250.
Question 27
Question
The population of a city is modeled by the exponential function P(t) = 5000 ·
1.02t, where trepresents the number of years since 2020. What will the popu-
lation be in the year 2030? Round your answer to the nearest whole number.
17
Solution
Step 1: Substitute t= 10 into the function to find the population in the year
2030.
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210.
1.0210 ≈1.218994
Step 3: Substitute the value of 1.0210 back into the equation.
P(10) = 5000 ·1.218994
Step 4: Calculate 5000 ·1.218994.
5000 ·1.218994 ≈6094
Therefore, the population of the city in the year 2030 will be approximately
6094.
Question 28
Question
A certain radioactive substance decays according to the formula A=A0e−kt,
where Ais the amount of substance remaining after tyears, A0is the initial
amount of substance, and kis a constant. If 80% of the substance decays in 30
years, what is the half-life of the substance?
Solution
Step 1: To find the half-life of the substance, we first need to determine the
value of kin the exponential decay formula A=A0e−kt .
Let A0be the initial amount of substance and Abe the amount remaining
after tyears. We are given that 80% of the substance decays in 30 years, so this
means that after 30 years, only 20% of the substance is left.
Step 2: Since only 20% of the substance remains after 30 years, we have:
0.20A0=A0e−30k
Step 3: Simplifying the equation, we find:
e−30k= 0.20
Step 4: Taking the natural logarithm of both sides gives:
lne−30k= ln(0.20)
18
Step 5: Using the property of logarithms that ln(ex) = x, we get:
−30k= ln(0.20)
Step 6: Solving for k, we find:
k=ln(0.20)
−30 ≈0.0231
Step 7: Now that we have the value of k, we can find the half-life of the
substance. The half-life is the time required for half of the substance to decay.
Step 8: Let A=1
2A0(half of the initial amount). We can plug this into the
exponential decay formula to find the half-life:
1
2A0=A0e−0.0231t
Step 9: Dividing by A0and solving for t, we get:
e0.0231t= 2
Step 10: Taking the natural logarithm of both sides gives:
lne0.0231t= ln(2)
Step 11: Simplifying, we find:
0.0231t= ln(2)
Step 12: Solving for t, we get:
t=ln(2)
0.0231 ≈30.03 years
Therefore, the half-life of the substance is approximately 30.03 years.
Question 29
Question
Given the exponential function P(t) = 200e0.03t, where trepresents time in
years, determine the time it takes for an initial investment of 200togrowto220.
Round your answer to the nearest hundredth.
Solution
Step 1: Set up the equation P(t) = 220 and solve for t.
200e0.03t= 220
e0.03t=220
200
e0.03t= 1.1
19
Step 2: Take the natural logarithm of both sides to solve for t.
lne0.03t= ln(1.1)
0.03t= ln(1.1)
t=ln(1.1)
0.03
Step 3: Use a calculator to find the approximate value of trounded to the
nearest hundredth.
t≈ln(1.1)
0.03
t≈0.09531
0.03
t≈3.18
Therefore, it takes approximately 3.18 years for the initial investment of
200togrowto220.
Question 30
Question
Solve the exponential equation 32x−1−5·3x+ 6 = 0 for x.
Solution
Step 1: Let’s rewrite the exponential equation as a quadratic equation:
(3x)2−5·3x+ 6 = 0
Step 2: Let y= 3x. Then the quadratic equation becomes:
y2−5y+ 6 = 0
Step 3: Now, let’s factor the quadratic equation:
(y−2)(y−3) = 0
Step 4: Set each factor to zero and solve for y:
y−2 = 0 =⇒y= 2 or y−3 = 0 =⇒y= 3
Step 5: Now that we have the possible values for y, let’s substitute back to
solve for x: For y= 2:
3x= 2
⇒x= log3(2)
For y= 3:
3x= 3
⇒x= 1
Step 6: Therefore, the solutions to the exponential equation are x= log3(2)
and x= 1.
20
Question 31
Question
Suppose the population of a city grows exponentially at a rate of 3.5% per year.
If the current population is 100,000, what will the population be in 10 years?
Solution
Step 1: To find the population in 10 years, we can use the formula for exponential
growth:
P(t) = P0×(1 + r)t
where: - P(t)is the population after time t, - P0is the initial population, - ris
the growth rate, and - tis the time period.
Step 2: Substituting the given values into the formula, we have:
P(10) = 100,000 ×(1 + 0.035)10
Step 3: Calculating the population after 10 years:
P(10) = 100,000 ×(1.035)10
P(10) = 100,000 ×1.418519
P(10) = 141,851.9
Step 4: Therefore, the population of the city will be approximately 141,852
in 10 years.
Question 32
Question
A company has determined that its monthly profit can be modeled by the func-
tion P(x) = 5000(1.10)x−5000, where xrepresents the number of months since
the company began operating. Find the number of months it will take for the
company to reach a monthly profit of 30,000.
Solution
Step 1: Set up the equation using the given information. We want to find the
number of months xwhen the monthly profit P(x)is 30,000.T hus, wehave :
30,000 = 5000(1.10)x−5000
Step 2: Solve for xby isolating the exponential term. We first add 5000 to
both sides of the equation:
35,000 = 5000(1.10)x
21
Step 3: Divide both sides by 5000 to isolate the exponential term:
7 = 1.10x
Step 4: Take the logarithm of both sides to solve for x. Let’s take the log1.10
of both sides:
log1.10 7 = log1.10 1.10x
Step 5: By the logarithm properties, the exponent can be moved down as a
coefficient:
log1.10 7 = x
Step 6: Finally, calculate the value of x. Using a calculator, we find:
x≈5.94
Therefore, it will take approximately 5.94 months for the company to reach
a monthly profit of 30,000.
Question 33
Question
Let f(x)=3·2xand g(x) = log2(x). Find the values of xthat satisfy f(x)>
g(x).
Solution
Step 1: We start by setting up the inequality f(x)> g(x)and using the given
functions f(x)and g(x):
3·2x>log2(x)
Step 2: To simplify the inequality, we can rewrite log2(x)as 2log2(x)=x:
3·2x> x
Step 3: Next, we can rewrite 3as 3·20and then simplify the inequality
further:
3·2x−0> x
2x+1 > x
Step 4: To solve this inequality, we can set up a graph for y= 2x+1 and
y=xon the same set of axes:
Step 5: By analyzing the graph, we see that the two functions intersect at
x= 1.
Step 6: To the determine the values of xfor which f(x)> g(x), we need to
consider the regions where f(x)is greater than g(x):
In the interval (−∞,1),2x+1 > x is true.
In the interval (1,∞),2x+1 < x is true.
Step 7: Therefore, the values of xthat satisfy f(x)> g(x)are x∈(−∞,1).
22
Question 34
Question
A certain radioactive substance decays according to the formula Q(t) = Q0·
e−0.025t, where Q(t)represents the amount of the substance remaining after t
years and Q0is the initial amount present. If 500 grams of the substance are
initially present, how many grams will be left after 10 years?
Solution
Step 1: Given that the initial amount Q0= 500 grams and the formula for the
decay is Q(t) = Q0·e−0.025t, we can substitute these values into the formula:
Q(t) = 500 ·e−0.025t
Step 2: To find the amount of the substance remaining after 10 years, we
substitute t= 10 into the formula:
Q(10) = 500 ·e−0.025·10
Step 3: Calculate the value of Q(10):
Q(10) = 500 ·e−0.25
Step 4: Evaluate e−0.25 using a calculator:
Q(10) = 500 ·0.77880078307
Step 5: Multiply to find the amount remaining after 10 years:
Q(10) ≈389.400391535
Therefore, after 10 years, there will be approximately 389.4 grams of the
substance remaining.
Question 35
Question
Solve the exponential equation 3x= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
23
Question 2
Question
The population of a city is modeled by the function P(t) = 12000 ·e0.02t, where
trepresents the number of years since the population was recorded. Find the
population of the city after 10 years, and determine the rate at which the pop-
ulation is growing at that time.
Solution
Step 1: To find the population of the city after 10 years, we substitute t= 10
into the population function P(t).
P(10) = 12000 ·e0.02·10
Step 2: Simplify the expression by calculating the exponent.
P(10) = 12000 ·e0.2
Step 3: Calculate the value of e0.2.
e0.2≈1.2214
Step 4: Multiply to find the population after 10 years.
P(10) ≈12000 ·1.2214 ≈14656.8
Therefore, the population of the city after 10 years is approximately 14,656.8.
Step 5: To determine the rate at which the population is growing after
10 years, we need to find the derivative of the population function P(t)with
respect to t, which gives us the instantaneous rate of change of the population
with respect to time. dP
dt = 240e0.02t
Step 6: Substitute t= 10 into dP
dt to find the rate of population growth after
10 years.
dP
dt
t=10
= 240e0.02·10
Step 7: Simplify and calculate the rate.
dP
dt
t=10
= 240e0.2≈293.1
Therefore, the rate at which the population is growing after 10 years is
approximately 293.1 individuals per year.
2
Question 3
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 original equation and solve for x.
32x+1 = 32
2x+ 1 = 2
Step 3: Solve for x.
2x= 1
x=1
2
Therefore, the solution to the exponential equation 32x+1 = 9 is x=1
2.
Question 4
Question
Solve for xin the equation 3x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the equation 3x−1= 27.
3x−1= 33
Step 3: Since the bases are the same, equate the exponents.
x−1 = 3
Step 4: Solve for xby adding 1to both sides of the equation.
x= 3 + 1
Step 5: Simplify the expression to find the value of x.
x= 4
Therefore, the solution to the equation 3x−1= 27 is x= 4.
3
Question 5
Question
Solve the following exponential equation for x:32x+1 = 5x−3.
Solution
Step 1: Take the natural logarithm of both sides of the equation to eliminate
the exponents:
ln32x+1= ln5x−3
Step 2: Apply the power rule of logarithms to simplify the left side of the
equation:
ln32x+1= (2x+ 1) ln(3)
Step 3: Similarly, apply the power rule of logarithms to simplify the right
side of the equation:
ln5x−3= (x−3) ln(5)
Step 4: Substitute back into the original equation to get:
(2x+ 1) ln(3) = (x−3) ln(5)
Step 5: Expand both sides of the equation:
2xln(3) + ln(3) = xln(5) −3 ln(5)
Step 6: Rearrange the equation to get all terms involving xon one side:
2xln(3) −xln(5) = −3 ln(5) −ln(3)
Step 7: Factor out the x:
x(2 ln(3) −ln(5)) = −3 ln(5) −ln(3)
Step 8: Solve for xby dividing both sides by (2 ln(3) −ln(5)):
x=−3 ln(5) −ln(3)
2 ln(3) −ln(5)
Therefore, the solution to the equation 32x+1 = 5x−3is x=−3 ln(5)−ln(3)
2 ln(3)−ln(5) .
Question 6
Question
Solve the exponential equation 4x−1= 32.
4
Solution
Step 1: Rewrite 32 as a power of 4.
32 = 42
Step 2: Substitute 42for 32 in the original equation.
4x−1= 42
Step 3: Set the exponents equal to each other.
x−1 = 2
Step 4: Solve for x.
x= 2 + 1
x= 3
Step 5: Check the solution.
Substitute x= 3 back into the original equation.
43−1= 32
42= 32
16 = 32
Since 16 = 32, the solution x= 3 is extraneous and the equation has no
solution.
Question 7
Question
Solve the exponential equation log2(x+ 3) −log2(x−3) = 3 for x.
Solution
Step 1: We start by using the properties of logarithms to simplify the equation.
Step 2: The difference of two logarithms with the same base is equal to the
logarithm of the division of their arguments. Thus, we have log2x+3
x−3= 3.
Step 3: Since loga(b) = ccan be rewritten as ac=b, we get 23=x+3
x−3. Step 4:
Simplifying 23gives us 8, so the equation becomes 8 = x+3
x−3. Step 5: Next, we
solve for xby multiplying both sides by x−3to get rid of the denominator. This
gives us 8(x−3) = x+3. Step 6: Expanding the left side, we get 8x−24 = x+3.
Step 7: Combining like terms, we have 7x= 27. Step 8: Finally, we solve for x
by dividing both sides by 7, giving us x=27
7.
Question 8
Question
Samantha invests $5000 in an account with an annual interest rate of 4% that
is compounded continuously. How much will be in the account after 10 years?
5
Solution
Step 1: We can use the formula for compound interest with continuous com-
pounding:
A=P·ert
where: - Ais the amount of money accumulated after tyears, including interest.
-Pis the principal amount (the initial amount of money). - ris the annual
interest rate (in decimal form). - tis the time the money is invested for (in
years). - e≈2.71828 is Euler’s number.
Step 2: In this case: - P= $5000 -r= 0.04 -t= 10
Step 3: Substituting the values into the formula:
A= 5000 ·e0.04·10
Step 4: Calculating the exponent:
A= 5000 ·e0.4
Step 5: Evaluating e0.4:
A= 5000 ·1.49182469
Step 6: Calculating the final amount:
A≈7459.12
Therefore, after 10 years, there will be approximately $7459.12 in the ac-
count.
Question 9
Question
Solve for xin the equation 2x+ 3 ·2x−1= 40.
Solution
Step 1: We can rewrite 3·2x−1as 3·2·2x−1= 6 ·2x−1.
Step 2: Substituting this back into the original equation gives us 2x+ 6 ·
2x−1= 40.
Step 3: Factor out 2x−1from both terms to simplify the equation:
2x−1(2 + 6) = 40
Step 4: Simplify the equation further to get:
2x−1·8 = 40
6
Step 5: Divide both sides by 8 to solve for 2x−1:
2x−1= 5
Step 6: To solve for x, we can rewrite 5as 22·23:
2x−1= 22·23
Step 7: Using the properties of exponents, we can simplify the equation to:
2x−1= 2x+2
Step 8: Setting the exponents equal to each other, we have:
x−1 = x+ 2
Step 9: Solving the equation for x, we get:
−1 = 2
Step 10: Since −1= 2, there is no solution to the equation 2x+3·2x−1= 40.
Question 10
Question
Solve the following exponential equation for x:
52x+1 = 25
Solution
Step 1: Rewrite both sides of the equation with the same base.
52x+1 = 25
52x+1 = 52
Step 2: Set the exponents equal to each other.
2x+ 1 = 2
Step 3: Solve for x.
2x+ 1 = 2
2x= 1
x=1
2
7
Step 4: Check the solution by substituting x=1
2back into the original
equation.
52( 1
2)+1 = 25
52+1 = 25
53= 25
125 = 25
Since the final step leads to a false statement, there is no solution to the
equation.
Question 11
Question
Solve the exponential equation 32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 as 33in the given equation.
32x−1= 33
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
2x−1 = 3
Step 4: Solve for x.
2x−1 = 3
2x= 4
x= 2
Step 5: Check the solution by substituting x= 2 back into the original
equation.
32(2)−1= 27
34−1= 33
33= 27
Therefore, the solution to the exponential equation 32x−1= 27 is x= 2.
8
Question 12
Question
The population of a certain city is modeled by the function P(t) = 10,000·e0.02t,
where tis the number of years since the start of 2020. Find and interpret the
population in the year 2030.
Solution
Step 1: To find the population in the year 2030, we need to substitute t=
2030 −2020 = 10 into the population function P(t).
P(10) = 10,000 ·e0.02·10
Step 2: Calculate the value of P(10).
P(10) = 10,000 ·e0.2
P(10) = 10,000 ·e0.2
P(10) ≈12214.29
Step 3: Interpretation: The population of the city in the year 2030 is ap-
proximately 12,214.29 people.
Question 13
Question
Solve for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Step 2: Solve for xusing the properties of
exponents and logarithms.
Step 1: Rewrite 8as a power of 2.
8 = 23
Step 2: Solve for xusing the properties of exponents and logarithms.
23x−1= 23
Since the bases are the same, we can equate the exponents:
3x−1 = 3
Solving for x:
3x= 4
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
9
Question 14
Question
You have just won a lottery that promises to pay you 80,000ayearf orthenext10years.However, youhavetheoptiontotakealumpsumpaymentnowinstead.T helumpsumpaymentisdeterminedusinganannualinterestrateof 5%.W hatistheminimumlumpsumamountyoushouldacceptnowif youcaninvestthemoneyat5%interestcompoundedannually?
Solution
Step 1: Calculate the present value of the annuity using the formula for the
present value of an annuity:
P V =P×1−1
(1 + r)n/r
Where, - P V is the present value of the annuity (the lump sum payment), -
Pis the annual payment amount (80,000),−ris the annual interest rate (5 or
0.05), - nis the number of years the annuity will be paid (10 years).
Plugging in the values, we get:
P V = 80000 ×1−1
(1 + 0.05)10 /0.05
Step 2: Calculate the present value of the annuity:
P V = 80000 ×1−1
1.0510 /0.05
P V = 80000 ×1−1
1.62889462677/0.05
P V = 80000 ×(1 −0.613913253) /0.05
P V = 80000 ×0.386086747/0.05
P V = 30887.39976
Therefore, the present value of the annuity, or the minimum lump sum
amount you should accept now, is $30,887.40.
Question 15
Question
Suppose the population of a city is modeled by the function P(t) = 5000 ·1.03t,
where trepresents the number of years after 2020. Find the year when the
population is expected to reach 8000.
10
Solution
Step 1: Set up the equation based on the information given. We want to find
the year when the population reaches 8000, so we set P(t) = 8000:
5000 ·1.03t= 8000
Step 2: Solve for tby dividing both sides by 5000:
1.03t=8000
5000
1.03t= 1.6
Step 3: Take the natural logarithm of both sides to solve for t:
ln1.03t= ln(1.6)
tln(1.03) = ln(1.6)
Step 4: Solve for tby dividing both sides by ln(1.03):
t=ln(1.6)
ln(1.03)
Step 5: Use a calculator to find the approximate value of t:
t≈ln(1.6)
ln(1.03) ≈12.52
Step 6: Add tto the base year 2020 to find the year when the population is
expected to reach 8000:
2020 + 12.52 ≈2032.52
Therefore, the population is expected to reach 8000 around the year 2033.
Question 16
Question
Solve the exponential equation 3x−2−3x−1= 8.
Solution
Step 1: Let’s rewrite the equation using the properties of exponents. We know
that am·an=am+n. So we can rewrite the given equation as 3x−2·3−1= 8.
Step 2: Simplifying the right side, we get 3x−1= 8.
Step 3: Now we can rewrite 8 as a power of 3. Since 32= 9, we have
8 = 3log38= 3 log 8
log 3 = 3 3
log 3 .
11
Step 4: Substituting the value of 8 back into our equation, we get 3x−1=
33
log 3 .
Step 5: Since the bases are the same, we can set the exponents equal to each
other. Thus, x−1 = 3
log 3 .
Step 6: To solve for x, we add 1 to both sides of the equation, giving us
x= 1 + 3
log 3 . Hence, the solution to the exponential equation is x= 1 + 3
log 3 .
Question 17
Question
Solve the exponential equation 2x+3 −2x+2 = 24.
Solution
Step 1: Rewrite the equation using the properties of exponents. Step 2: Apply
the properties of exponents to simplify the equation. Step 3: Solve for xby
isolating 2x. Step 4: Check the solution to ensure it is valid.
Step 1: Rewrite the equation using the properties of exponents.
We rewrite 2x+3 −2x+2 = 24 as 2·2x−2·2x= 24.
Step 2: Apply the properties of exponents to simplify the equation.
Simplifying further, we get 2x−2x= 24.
Step 3: Solve for xby isolating 2x.
Since 2x−2x= 0, we have 0 = 24 which is not possible. Therefore, there is
no solution to the equation 2x+3 −2x+2 = 24.
Question 18
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 exponents equal to each other.
2x+ 1 = 3
Step 3: Solve for x.
2x+ 1 = 3
2x= 2
x= 1
12
Step 4: Check the solution.
32(1)+1 = 33= 27
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
Question 19
Question
Solve the following exponential equation for x:23x= 8.
Solution
Step 1: Rewrite 8 as a power of 2. Step 2: Solve for xby equating the exponents.
Question 20
Question
Solve for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3. We know that 33= 27, so 27 = 33.
Step 2: Substitute 27 with 33in the equation 32x−1= 27. We get 32x−1= 33.
Step 3: Since the bases are equal, we can set the exponents equal to each
other. So, 2x−1 = 3.
Step 4: Add 1 to both sides of the equation to isolate the term with x:
2x= 4.
Step 5: Divide by 2 on both sides to solve for x:x= 2.
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 21
Question
Solve the logarithmic equation: log5(3x+ 1) + log5(x−2) = 2.
Solution
Step 1: Apply the product rule of logarithms to combine the logarithms on the
left side.
log5[(3x+ 1)(x−2)] = 2
13
Step 2: Rewrite the equation in exponential form. Remember that loga(b) =
cis equivalent to ac=b.
52= (3x+ 1)(x−2)
Step 3: Simplify the equation by squaring 5.
25 = 3x2−6x+x−2
Step 4: Combine like terms and set the equation equal to zero.
3x2−5x−27 = 0
Step 5: Factor the quadratic equation.
(3x+ 9)(x−3) = 0
Step 6: Solve for x by setting each factor to zero.
3x+ 9 = 0 or x−3 = 0
Step 7: Solve for x in each equation to find the possible values.
x=−3or x= 3
Step 8: Check for extraneous solutions by substituting the values back into
the original logarithmic equation.
x=−3 : log5(3(−3) + 1) + log5((−3) −2) = log5(−8) + undefined
x= 3 : log5(3(3) + 1) + log5(3 −2) = log5(10) + log5(1) = 1 + 0 = 2
Step 9: Therefore, the solution to the logarithmic equation is x= 3.
Question 22
Question
Suppose a bacteria culture starts with 100 bacteria and grows at a rate of 8%
per hour. Write an exponential growth model for the number of bacteria after
thours and determine how many bacteria will be present after 5 hours.
Solution
Step 1: Let N(t)represent the number of bacteria after thours. Since the pop-
ulation is growing at a rate of 8% per hour, the growth model can be expressed
as:
N(t) = 100 ×(1 + 0.08)t
Step 2: Simplify the model to get a final formula for N(t):
N(t) = 100 ×(1.08)t
14
Step 3: To find the number of bacteria after 5 hours, substitute t= 5 into
the formula:
N(5) = 100 ×(1.08)5
N(5) = 100 ×1.46933
N(5) ≈146.93
Therefore, there will be approximately 146.93 bacteria present after 5 hours.
Question 23
Question
Suppose the population of a town is modeled by the function P(t) = 5000·e0.02t,
where trepresents the number of years since the initial population count was
taken. Find the population after 10 years to the nearest whole number.
Solution
Step 1: Substitute t= 10 into the given function to find the population after
10 years.
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression inside the exponent.
P(10) = 5000 ·e0.2
Step 3: Calculate the exponential term.
P(10) = 5000 ·1.221402
Step 4: Multiply to find the population after 10 years.
P(10) ≈6107
Therefore, the population of the town after 10 years is approximately 6107
people.
Question 24
Question
Solve the exponential equation 32x+1 −2·3x= 5 for x.
15
Solution
Step 1: Let’s rewrite the given equation into a more manageable form by using
a substitution. Let u= 3x. Then, the equation becomes 3u2−2u= 5.
Step 2: Rearrange the equation 3u2−2u= 5 into a quadratic form by
moving all terms to one side of the equation. We get 3u2−2u−5 = 0.
Step 3: Now, we can solve the quadratic equation 3u2−2u−5=0for u
using the quadratic formula: u=−(−2)±√(−2)2−4·3·(−5)
2·3.
Step 4: Simplify the quadratic formula to find the solutions for u:u=
2±√4+60
6. Thus, u=2±√64
6.
Step 5: Continuing to simplify, we get two solutions: u=2±8
6. So, u=10
6
or u=−6
6.
Step 6: Therefore, u=5
3or u=−1. Recall that u= 3x, so we have 3x=5
3
or 3x=−1. However, 3xcannot be negative, so we focus on the equation
3x=5
3.
Step 7: To solve 3x=5
3, take the natural logarithm of both sides: ln(3x) =
ln 5
3.
Step 8: Apply the logarithmic property to bring down the exponent: xln(3) =
ln 5
3.
Step 9: Solve for x:x=ln(5
3)
ln(3) . This is the exact solution for x.
Question 25
Question
Suppose a bacteria population doubles every 6 hours. If there are initially 100
bacteria, how many bacteria will there be after 24 hours?
Solution
Step 1: Find the growth factor from the doubling time.
Given that the population doubles every 6 hours, we can find the growth
factor, r, using the formula:
r= 2 1
doubling time = 2 1
6
Step 2: Calculate the growth factor.
r= 21
6= 21
6= 21
6= 20.1667 ≈1.1225
Step 3: Use the growth factor to find the population after 24 hours.
With an initial population of 100 bacteria and a growth factor of 1.1225, the
population after 24 hours can be calculated using the formula for exponential
growth:
Population after 24 hours =Initial population ×(1.1225) time (in hours)
doubling time
16
Population after 24 hours = 100 ×(1.1225)24
6
Population after 24 hours = 100 ×(1.1225)4
Population after 24 hours = 100 ×(1.601)4= 100 ×2.564 ≈256.4
Therefore, after 24 hours, there will be approximately 256 bacteria in the
population.
Question 26
Question
The population of a city is modeled by the function P(t) = 50,000·1.02t, where
trepresents the number of years since 2020. Calculate the population of the
city in the year 2040.
Solution
Step 1: Determine the number of years from 2020 to 2040. We need to find t
when the year is 2040.
2040 −2020 = 20 years
Step 2: Substitute the value of tinto the population function.
P(20) = 50,000 ·1.0220
Step 3: Calculate 1.0220.
1.0220 ≈1.485
Step 4: Substitute this value back into the population function.
P(20) = 50,000 ·1.485
P(20) = 74250
So, the population of the city in the year 2040 will be 74,250.
Question 27
Question
The population of a city is modeled by the exponential function P(t) = 5000 ·
1.02t, where trepresents the number of years since 2020. What will the popu-
lation be in the year 2030? Round your answer to the nearest whole number.
17
Solution
Step 1: Substitute t= 10 into the function to find the population in the year
2030.
P(10) = 5000 ·1.0210
Step 2: Calculate 1.0210.
1.0210 ≈1.218994
Step 3: Substitute the value of 1.0210 back into the equation.
P(10) = 5000 ·1.218994
Step 4: Calculate 5000 ·1.218994.
5000 ·1.218994 ≈6094
Therefore, the population of the city in the year 2030 will be approximately
6094.
Question 28
Question
A certain radioactive substance decays according to the formula A=A0e−kt,
where Ais the amount of substance remaining after tyears, A0is the initial
amount of substance, and kis a constant. If 80% of the substance decays in 30
years, what is the half-life of the substance?
Solution
Step 1: To find the half-life of the substance, we first need to determine the
value of kin the exponential decay formula A=A0e−kt .
Let A0be the initial amount of substance and Abe the amount remaining
after tyears. We are given that 80% of the substance decays in 30 years, so this
means that after 30 years, only 20% of the substance is left.
Step 2: Since only 20% of the substance remains after 30 years, we have:
0.20A0=A0e−30k
Step 3: Simplifying the equation, we find:
e−30k= 0.20
Step 4: Taking the natural logarithm of both sides gives:
lne−30k= ln(0.20)
18
Step 5: Using the property of logarithms that ln(ex) = x, we get:
−30k= ln(0.20)
Step 6: Solving for k, we find:
k=ln(0.20)
−30 ≈0.0231
Step 7: Now that we have the value of k, we can find the half-life of the
substance. The half-life is the time required for half of the substance to decay.
Step 8: Let A=1
2A0(half of the initial amount). We can plug this into the
exponential decay formula to find the half-life:
1
2A0=A0e−0.0231t
Step 9: Dividing by A0and solving for t, we get:
e0.0231t= 2
Step 10: Taking the natural logarithm of both sides gives:
lne0.0231t= ln(2)
Step 11: Simplifying, we find:
0.0231t= ln(2)
Step 12: Solving for t, we get:
t=ln(2)
0.0231 ≈30.03 years
Therefore, the half-life of the substance is approximately 30.03 years.
Question 29
Question
Given the exponential function P(t) = 200e0.03t, where trepresents time in
years, determine the time it takes for an initial investment of 200togrowto220.
Round your answer to the nearest hundredth.
Solution
Step 1: Set up the equation P(t) = 220 and solve for t.
200e0.03t= 220
e0.03t=220
200
e0.03t= 1.1
19
Step 2: Take the natural logarithm of both sides to solve for t.
lne0.03t= ln(1.1)
0.03t= ln(1.1)
t=ln(1.1)
0.03
Step 3: Use a calculator to find the approximate value of trounded to the
nearest hundredth.
t≈ln(1.1)
0.03
t≈0.09531
0.03
t≈3.18
Therefore, it takes approximately 3.18 years for the initial investment of
200togrowto220.
Question 30
Question
Solve the exponential equation 32x−1−5·3x+ 6 = 0 for x.
Solution
Step 1: Let’s rewrite the exponential equation as a quadratic equation:
(3x)2−5·3x+ 6 = 0
Step 2: Let y= 3x. Then the quadratic equation becomes:
y2−5y+ 6 = 0
Step 3: Now, let’s factor the quadratic equation:
(y−2)(y−3) = 0
Step 4: Set each factor to zero and solve for y:
y−2 = 0 =⇒y= 2 or y−3 = 0 =⇒y= 3
Step 5: Now that we have the possible values for y, let’s substitute back to
solve for x: For y= 2:
3x= 2
⇒x= log3(2)
For y= 3:
3x= 3
⇒x= 1
Step 6: Therefore, the solutions to the exponential equation are x= log3(2)
and x= 1.
20
Question 31
Question
Suppose the population of a city grows exponentially at a rate of 3.5% per year.
If the current population is 100,000, what will the population be in 10 years?
Solution
Step 1: To find the population in 10 years, we can use the formula for exponential
growth:
P(t) = P0×(1 + r)t
where: - P(t)is the population after time t, - P0is the initial population, - ris
the growth rate, and - tis the time period.
Step 2: Substituting the given values into the formula, we have:
P(10) = 100,000 ×(1 + 0.035)10
Step 3: Calculating the population after 10 years:
P(10) = 100,000 ×(1.035)10
P(10) = 100,000 ×1.418519
P(10) = 141,851.9
Step 4: Therefore, the population of the city will be approximately 141,852
in 10 years.
Question 32
Question
A company has determined that its monthly profit can be modeled by the func-
tion P(x) = 5000(1.10)x−5000, where xrepresents the number of months since
the company began operating. Find the number of months it will take for the
company to reach a monthly profit of 30,000.
Solution
Step 1: Set up the equation using the given information. We want to find the
number of months xwhen the monthly profit P(x)is 30,000.T hus, wehave :
30,000 = 5000(1.10)x−5000
Step 2: Solve for xby isolating the exponential term. We first add 5000 to
both sides of the equation:
35,000 = 5000(1.10)x
21
Step 3: Divide both sides by 5000 to isolate the exponential term:
7 = 1.10x
Step 4: Take the logarithm of both sides to solve for x. Let’s take the log1.10
of both sides:
log1.10 7 = log1.10 1.10x
Step 5: By the logarithm properties, the exponent can be moved down as a
coefficient:
log1.10 7 = x
Step 6: Finally, calculate the value of x. Using a calculator, we find:
x≈5.94
Therefore, it will take approximately 5.94 months for the company to reach
a monthly profit of 30,000.
Question 33
Question
Let f(x)=3·2xand g(x) = log2(x). Find the values of xthat satisfy f(x)>
g(x).
Solution
Step 1: We start by setting up the inequality f(x)> g(x)and using the given
functions f(x)and g(x):
3·2x>log2(x)
Step 2: To simplify the inequality, we can rewrite log2(x)as 2log2(x)=x:
3·2x> x
Step 3: Next, we can rewrite 3as 3·20and then simplify the inequality
further:
3·2x−0> x
2x+1 > x
Step 4: To solve this inequality, we can set up a graph for y= 2x+1 and
y=xon the same set of axes:
Step 5: By analyzing the graph, we see that the two functions intersect at
x= 1.
Step 6: To the determine the values of xfor which f(x)> g(x), we need to
consider the regions where f(x)is greater than g(x):
In the interval (−∞,1),2x+1 > x is true.
In the interval (1,∞),2x+1 < x is true.
Step 7: Therefore, the values of xthat satisfy f(x)> g(x)are x∈(−∞,1).
22
Question 34
Question
A certain radioactive substance decays according to the formula Q(t) = Q0·
e−0.025t, where Q(t)represents the amount of the substance remaining after t
years and Q0is the initial amount present. If 500 grams of the substance are
initially present, how many grams will be left after 10 years?
Solution
Step 1: Given that the initial amount Q0= 500 grams and the formula for the
decay is Q(t) = Q0·e−0.025t, we can substitute these values into the formula:
Q(t) = 500 ·e−0.025t
Step 2: To find the amount of the substance remaining after 10 years, we
substitute t= 10 into the formula:
Q(10) = 500 ·e−0.025·10
Step 3: Calculate the value of Q(10):
Q(10) = 500 ·e−0.25
Step 4: Evaluate e−0.25 using a calculator:
Q(10) = 500 ·0.77880078307
Step 5: Multiply to find the amount remaining after 10 years:
Q(10) ≈389.400391535
Therefore, after 10 years, there will be approximately 389.4 grams of the
substance remaining.
Question 35
Question
Solve the exponential equation 3x= 27.
Solution
Step 1: Rewrite 27 as a power of 3:
27 = 33
23
Step 2: Substitute 33for 27 in the equation 3x= 27:
3x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
x= 3
Therefore, the solution to the exponential equation 3x= 27 is x= 3.
24
