MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 7
Liberty University
Question 1
Question
Suppose the population of a city follows an exponential growth model given by
P(t) = 5000 ·1.03t, where P(t)represents the population after tyears. In how
many years will the population reach 10,000?
Solution
Step 1: Set up the equation using the given information. We are given that
the population after tyears is given by P(t) = 5000 ·1.03t. We want to find
the value of twhen the population reaches 10,000. So, we set up the equation
5000 ·1.03t= 10000.
Step 2: Solve the equation for t. Divide both sides by 5000 to isolate the
exponential term:
1.03t=10000
5000 = 2
Step 3: Take the natural logarithm of both sides.
ln(1.03t)= ln(2)
tln(1.03) = ln(2)
Step 4: Solve for t.
t=ln(2)
ln(1.03)
t≈0.6931
0.0296
t≈23.42
Step 5: Interpret the solution. It will take approximately 23.42 years for the
population of the city to reach 10,000. This means that after approximately
23.42 years, the city’s population will have doubled from its initial population
of 5,000 people.
Question 2
Question
Given that f(x) = 3xand g(x) = log3(x), find the value of (f◦g)(9).
Solution
Step 1: First, we find g(9) by substituting x= 9 into the function g(x) = log3(x).
g(9) = log3(9)
Step 2: We know that 32= 9, so log3(9) = 2.
g(9) = 2
Step 3: Next, we find (f◦g)(9) by substituting g(9) = 2 into the function
f(x) = 3x.
(f◦g)(9) = f(g(9)) = f(2)
Step 4: Substituting x= 2 into the function f(x) = 3xgives:
f(2) = 32
Step 5: Therefore, (f◦g)(9) = 32= 9 .
Question 3
Question
Solve the following exponential equation: 3x−2·3x−1= 8.
Solution
Step 1: Rewrite the equation in terms of a single base. We can rewrite 3xas
3·3x−1.
Step 2: Substitute 3·3x−1back into the equation. 3x−2·3x−1= 3 ·3x−1−
2·3x−1= 3x−1
Step 3: Solve the simplified equation. 3x−1= 8
Step 4: Rewrite 8 as a power of 3. 3x−1= 3log38
Step 5: Set the exponents equal to each other. x−1 = log38
2
Step 6: Solve for x.x= log38+1
Step 7: Evaluate the expression. x= log38 + 1 = log323+ 1 = 3 log32 + 1 =
3 + 1 = 4
Therefore, the solution to the equation is x= 4.
Question 4
Question
Sara invests $4000 in a savings account that pays 3.5% interest compounded
continuously. How long will it take for Sara’s investment to double in value?
Solution
Step 1: The formula for continuously compounded interest is given by A=P ert,
where: - Ais the amount of money accumulated after tyears, including interest,
-Pis the principal amount (the initial amount of money), - ris the annual
interest rate (in decimal form), - tis the time the money is invested for in years,
and - eis the base of the natural logarithm, approximately equal to 2.71828.
Step 2: In this case, Sara wants her investment to double to $8000. So, we
have the following equation:
8000 = 4000e0.035t
Step 3: Divide by 4000 to isolate the exponential term:
2 = e0.035t
Step 4: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.035t)
Step 5: Use the property of logarithms ln(ex) = x:
ln 2 = 0.035t
Step 6: Solve for tby dividing by 0.035:
t=ln 2
0.035
Step 7: Calculate the value of tusing a calculator:
t≈0.69315
0.035 ≈19.80429
Therefore, it will take approximately 19.8 years for Sara’s investment to
double in value when compounded continuously at a rate of 3.5%.
3
Question 5
Question
Solve for xin the equation 2x−1−3·2x+ 2x+1 = 0.
Solution
Step 1: Rewrite the given equation using properties of exponents:
2x−1−3·2x+ 2x+1 = 0 ⇒2x
2−3·2x
1+ 2 ·2x= 0
Step 2: Find a common denominator to combine the terms:
2x
2−6·2x
2+4·2x
2= 0
Step 3: Combine the terms:
2x−6·2x+ 4 ·2x
2= 0 ⇒−3·2x
2= 0
Step 4: Solve for 2x:
−3·2x= 0 ⇒2x= 0
Since 2xcannot equal 0, there are no solutions to the equation 2x−1−3·
2x+ 2x+1 = 0.
Question 6
Question
Suppose a certain investment grows according to the formula A=P·e0.06t,
where Arepresents the final amount, Pis the principal investment, tis the
time in years, and eis Euler’s number. If an initial investment of $5000 grows
to be $8000, find the time it takes for the investment to double.
Solution
Let’s denote the time it takes for the investment to double as td. To find this
value, we need to set Aequal to 2 times the initial investment (2P) and solve
for td.
Step 1: Set up the equation using the given information. We know that
A= 2P= 2 ·5000 = 10000, so we have:
10000 = 5000 ·e0.06td
4
Step 2: Solve for tdby isolating the variable. Divide both sides by 5000:
e0.06td= 2
Step 3: Take the natural logarithm of both sides to solve for td.
ln(e0.06td)= ln(2)
0.06td= ln(2)
td=ln(2)
0.06
td≈0.693
0.06
td≈11.55
Therefore, it will take approximately 11.55 years for the investment to dou-
ble.
Question 7
Question
The population of a city is growing exponentially at a rate of 3.5% per year. If
the current population is 500,000, what will the population be in 10 years?
Solution
Step 1: Identify the given information and the formula for exponential growth.
Given: - Population growth rate: 3.5% per year - Current population: 500,000
We can model exponential growth using the formula:
P(t) = P0×(1 + r)t
where: - P(t)is the population after tyears, - P0is the initial population, - r
is the growth rate as a decimal, - tis the time in years.
Step 2: Convert the growth rate to a decimal. The growth rate is 3.5%,
which is equivalent to 0.035 as a decimal.
Step 3: Substitute the given values into the formula and solve for the pop-
ulation after 10 years. Plugging in the values:
P(10) = 500,000 ×(1 + 0.035)10
P(10) = 500,000 ×(1.035)10
Step 4: Calculate the population after 10 years.
P(10) = 500,000 ×1.4190675
P(10) ≈709,534
Therefore, the population of the city will be approximately 709,534 in 10
years.
5
Question 8
Question
Solve the following exponential equation for x:3x+1 −2·3x+ 1 = 0.
Solution
Step 1: Let u= 3x. Then the equation becomes u2−2u+ 1 = 0.
Step 2: Solve the quadratic equation u2−2u+ 1 = 0 by factoring:
(u−1)2= 0
Step 3: Set (u−1)2= 0 equal to zero and solve for u:
u−1 = 0 =⇒u= 1
Step 4: Substitute back u= 3xinto u= 1 to get:
3x= 1
Step 5: Solve 3x= 1 to find x:
x= 0
Therefore, the solution to the equation 3x+1 −2·3x+ 1 = 0 is x= 0.
Question 9
Question
Samantha invested $10,000 in a savings account that earns 4% annual interest
compounded quarterly. How long will it take for her investment to double in
value?
Solution
Step 1: The formula for compound interest is given by:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after tyears, including interest,
P= the principal amount (the initial amount of money), r= annual interest
rate (in decimal form), n= number of times that interest is compounded per
year, t= time the money is invested for in years.
Given the information: P= $10,000,r= 0.04,n= 4, Samantha wants to
find twhen A= 2P= 2 ·10000 = $20,000. Substituting these values into the
formula:
20000 = 10000 (1 + 0.04
4)4t
6
Step 2: Simplifying the equation to solve for t:
2 = (1 + 0.01)4t
Step 3: Rewrite the equation:
2 = (1.01)4t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln (1.01)4t
Step 5: Use the property of logarithms to bring down the exponent:
ln(2) = 4t·ln(1.01)
Step 6: Solve for t:
t=ln(2)
4 ln(1.01) ≈0.6931
4·0.0099 ≈17.82 years
Therefore, it will take approximately 17.82 years for Samantha’s investment
to double in value.
Question 10
Question
Samantha invests $5000 in a savings account that earns an annual interest rate of
4% compounded quarterly. Write an exponential function to model the amount
of money A(t)in the account after tyears.
Solution
Step 1: We know that the formula for compound interest is given by:
A(t) = P(1 + r
n)nt
where: - P= $5000 (initial investment) - r= 0.04 (annual interest rate) - n= 4
(number of times interest is compounded in a year) - tis the time in years
Step 2: Substitute the known values into the formula:
A(t) = 5000 (1 + 0.04
4)4t
Step 3: Simplify the expression inside the parentheses:
A(t) = 5000 (1 + 0.01)4t
Step 4: Further simplify the expression:
A(t) = 5000 (1.01)4t
Therefore, the exponential function to model the amount of money A(t)in
the account after tyears is A(t) = 5000(1.01)4t.
7
Question 11
Question
Solve the following exponential equation for x:23x−1−3·2x−1−2 = 0.
Solution
Step 1: Let’s substitute u= 2x−1to simplify the equation. This gives us:
2·u3−3u−2 = 0.
Step 2: Factor the left side of the equation to get: 2u3−4u+u−2=0⇒
2u(u2−2) + 1(u−2) = 0.
Step 3: Set each factor to zero and solve for u. We get u= 2 from u−2 = 0
and u=±√2from u2−2 = 0.
Step 4: Now substitute 2x−1back in for u. So, we have 2x−1= 2 and
2x−1=±√2.
Step 5: Solve the first equation 2x−1= 2 to find x. This simplifies to
x−1 = 1 ⇒x= 2.
Step 6: Solve the second equation 2x−1=±√2to find x. This simplifies to
x−1 = 1
2⇒x=3
2.
Step 7: Therefore, the solutions to the exponential equation 23x−1−3·2x−1−
2 = 0 are x= 2 and x=3
2.
Question 12
Question
Samantha invests $10,000 in an account that earns 4.5% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Step 1: To find the time it takes for an investment to double in value when
interest is compounded continuously, we can use the formula:
A=P·ert
Where: - Ais the amount of money after time t, - Pis the principal amount
(initial investment), - ris the annual interest rate, - tis the time the money is
invested for, and - eis the base of natural logarithms (approximately 2.71828).
Step 2: In this case, Samantha invests $10,000, so P= 10000, the interest
rate is 4.5% or 0.045, and she wants her investment to double, which means
A= 2P= 20000.
Step 3: Substituting the given values into the formula, we have:
20000 = 10000 ·e0.045t
8
Step 4: Divide both sides by 10000 to isolate the exponential expression:
2 = e0.045t
Step 5: To solve for t, we need to take the natural logarithm of both sides:
ln(2) = ln(e0.045t)
Step 6: Since ln(ex) = x, we can simplify our equation to:
ln(2) = 0.045t
Step 7: Now, divide both sides by 0.045 to solve for t:
t=ln(2)
0.045
Step 8: Using a calculator, we find:
t≈0.693147
0.045 ≈15.47
Step 9: Therefore, it will take approximately 15.47 years for Samantha’s
investment to double in value.
Question 13
Question
A certain radioactive substance decays according to the function Q(t) = Q0·
e−0.02t, where Q(t)represents the quantity of the substance remaining after t
years and Q0is the initial quantity of the substance. If the initial quantity of
the substance is 500 grams, find: a) The quantity of the substance that remains
after 10 years. b) The time it takes for the quantity of the substance to drop to
200 grams.
Solution
a) To find the quantity of the substance that remains after 10 years, we can
simply substitute t= 10 into the given exponential decay function.
Q(10) = 500 ·e−0.02·10
Q(10) = 500 ·e−0.2
Q(10) ≈500 ·0.8187
Q(10) ≈409.35 grams
9
b) To find the time it takes for the quantity of the substance to drop to 200
grams, we need to solve the equation:
200 = 500 ·e−0.02t
Dividing both sides by 500:
0.4 = e−0.02t
Taking the natural logarithm of both sides:
ln(0.4) = ln(e−0.02t)
ln(0.4) = −0.02t
Solving for t:
t=ln(0.4)
−0.02
t≈−0.9163
−0.02
t≈45.82 years
Therefore, it takes approximately 45.82 years for the quantity of the sub-
stance to drop to 200 grams.
Question 14
Question
A population of 1000 bacteria doubles in size every 12 hours. Write an expo-
nential growth model to represent the population Pof bacteria as a function of
time tin days. Round your answer to three decimal places.
Solution
Step 1: Since the population of bacteria doubles every 12 hours, the growth rate
is 100% every 12 hours, or 200% in one full day (24 hours). This translates to
a growth factor of 2.
Step 2: We can express the population Pas a function of time tin days
using the formula for exponential growth:
P(t) = P0·2rt,
where: - P0is the initial population, - ris the growth rate (in this case, 1
because the population doubles), - tis the time in days.
Step 3: Since the population starts at 1000 bacteria, we have P0= 1000.
Therefore, the exponential growth model for the population of bacteria is:
P(t) = 1000 ·21t.
10
Step 4: Simplifying the model gives:
P(t) = 1000 ·2t.
Step 5: Rounding the answer to three decimal places, the exponential growth
model for the population of bacteria as a function of time is:
P(t) = 1000 ·2t.
Question 15
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where
tis the time in years. Find the population of the city after 10 years.
Solution
Step 1: To find the population after 10 years, substitute t= 10 into the function
P(t).
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression inside the exponential function.
P(10) = 5000 ·e0.2
Step 3: Calculate the value of e0.2.
P(10) = 5000 ·e0.2= 5000 ·1.221402758
Step 4: Multiply the population by the calculated value of e0.2.
P(10) = 6107.01379
Therefore, the population of the city after 10 years is approximately 6,107
people.
Question 16
Question
A certain radioactive substance decays exponentially. If 100 milligrams of the
substance decays to 70 milligrams in 10 hours, how much of the substance will
remain after 24 hours?
11
Solution
Step 1: Determine the decay constant kusing the given information. Step
2: Use the exponential decay formula to calculate how much of the substance
remains after 24 hours.
Step 1: First, we have the exponential decay formula:
A(t) = A0·e−kt
where: - A(t)is the amount of substance remaining after thours, - A0is the
initial amount of the substance, - kis the decay constant, - eis Euler’s number
(approximately 2.71828).
Given that 100 milligrams decays to 70 milligrams in 10 hours, we can set
up the following equation:
70 = 100 ·e−10k
To solve for k, divide both sides by 100:
0.7 = e−10k
Taking the natural logarithm of both sides:
ln(0.7) = ln(e−10k)
ln(0.7) = −10k
Dividing by -10 to solve for k:
k=−ln(0.7)
10
Step 2: Now that we have found the decay constant k, we can use the
exponential decay formula to calculate how much of the substance remains after
24 hours:
A(24) = 100 ·e−k·24
Substitute k=−ln(0.7)
10 into the formula:
A(24) = 100 ·e−
ln(0.7)
10
·24
Calculating this expression will give us the amount of the substance remain-
ing after 24 hours.
Question 17
Question
Let f(x) = 2x+ 5 and g(x) = log3(x−1). Find the composite function f(g(x)).
12
Solution
Step 1: Find f(g(x)) by substituting g(x)into f(x). Step 2: Rewrite the
composite function in simplified form.
Step 1:
f(g(x)) = f(log3(x−1)) = 2log3(x−1) + 5
Step 2: To simplify 2log3(x−1), we can use the property alogab=b. Hence,
2log3(x−1) =x−1
Therefore, the composite function f(g(x)) simplifies to:
f(g(x)) = x−1 + 5 = x+ 4
So, f(g(x)) = x+ 4.
Question 18
Question
Solve for xin the equation 52x−1= 125.
Solution
Step 1: Rewrite 125 as a power of 5by observing that 125 = 53. Step 2:
Substitute 125 as 53in the equation 52x−1= 125 to get 52x−1= 53. Step 3:
Since the bases are the same, equate the exponents: 2x−1 = 3. Step 4: Solve
the equation 2x−1=3for xby first adding 1to both sides to get 2x= 4.
Step 5: Divide by 2on both sides to solve for x:x= 2. Step 6: Therefore, the
solution to the equation 52x−1= 125 is x= 2.
Question 19
Question
Solve the exponential equation 3x+1 = 27 for x.
Solution
Step 1: Rewrite the base 27 as a power of 3.
27 = 33
Step 2: Substitute 33for 27 in the original equation.
3x+1 = 33
13
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
x+ 1 = 3
Step 4: Solve for xby subtracting 1from both sides.
x= 3 −1
x= 2
Step 5: Therefore, the solution to the exponential equation 3x+1 = 27 is
x= 2.
Question 20
Question
A colony of bacteria doubles in size every 4 hours. If there are initially 100
bacteria in the colony, how many bacteria will there be after 24 hours?
Solution
Step 1: Determine the growth factor for the bacteria colony. Since the colony
doubles in size every 4 hours, the growth factor ris 2.
Step 2: Write the exponential growth model for the bacteria colony. Let
N(t)be the number of bacteria at time tin hours. The model is given by:
N(t) = 100 ×2t
4
Step 3: Substitute t= 24 into the model to find the number of bacteria after
24 hours.
N(24) = 100 ×224
4
Step 4: Simplify the expression.
N(24) = 100 ×26
N(24) = 100 ×64
N(24) = 6400
Step 5: Therefore, after 24 hours, there will be 6400 bacteria in the colony.
Question 21
Question
Solve the exponential equation for x:4x+1 = 32.
14
Solution
Step 1: Rewrite 32 as a power of 2. We have 32 = 25.
Step 2: Rewrite the given exponential equation with a common base, 4 and
2 in this case. 4x+1 = (22)x+1.
Step 3: Apply the power rule of exponents by multiplying the exponents
inside the parentheses. 4x+1 = 22x+2.
Step 4: Since 4x+1 = 22x+2, we can set the exponents equal to each other.
22x+2 = 25.
Step 5: Now equate the exponents. 2x+ 2 = 5.
Step 6: Subtract 2 from both sides of the equation. 2x= 3.
Step 7: Finally, divide by 2 to solve for x.x=3
2.
Therefore, the solution to the exponential equation 4x+1 = 32 is x=3
2.
Question 22
Question
For an investment that earns 5
Solution
Step 1: Let’s start by setting up the equation for when the investment doubles
in value. Since the initial investment is 1000, we want to find the time twhen
the value of the investment is 2000. Therefore, we have:
2000 = 1000(1.05)t
Step 2: Divide both sides by 1000 to isolate the exponential term:
2000
1000 = (1.05)t
Step 3: Simplify the left side of the equation:
2 = (1.05)t
Step 4: To solve for t, we take the natural logarithm (ln) of both sides:
ln(2) = ln((1.05)t)
Step 5: Use the property of logarithms that allows us to bring the exponent
down in front:
ln(2) = tln(1.05)
Step 6: Now, divide both sides by ln(1.05) to solve for t:
t=ln(2)
ln(1.05)
15
Step 7: Use a calculator to find the approximate value of t:
t≈ln(2)
ln(1.05) ≈0.6931
0.0488 ≈14.20
Step 8: Therefore, it will take approximately 14.20 years for the investment
to double in value.
Question 23
Question
The population of a city is currently 50,000 and is projected to double every 10
years. (a) Find the exponential function that models the population of the city
as a function of time in years. (b) Determine the population of the city after 30
years.
Solution
(a) Let P(t)represent the population of the city as a function of time tin
years. Since the population doubles every 10 years, the exponential function
that models the population growth is given by:
P(t) = 50000 ×2t/10
(b) To determine the population of the city after 30 years, we substitute
t= 30 into the exponential function:
P(30) = 50000 ×230/10
P(30) = 50000 ×23
P(30) = 50000 ×8
P(30) = 400000
Therefore, the population of the city after 30 years is 400,000.
Question 24
Question
Solve for x:32x−1= 27.
16
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,the equation becomes 32x−1= 33.
Step 2: Equate the exponents.
Since the bases are equal, we have 2x−1 = 3.
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2.
Step 4: Verify the solution.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
33= 33=⇒27 = 27.
Therefore, the solution is x= 2.
Question 25
Question
Solve the following exponential equation for x:52x−3= 25.
Solution
Step 1: Rewrite 25 as a power of 5.
25 = 52
Step 2: Substitute 52back into the original equation.
52x−3= 52
Step 3: Since the bases are the same, we can set the exponents equal to
each other.
2x−3 = 2
Step 4: Add 3to both sides of the equation.
2x= 5
Step 5: Divide by 2to solve for x.
x=5
2
Therefore, the solution to the equation 52x−3= 25 is x=5
2.
17
Question 26
Question
Solve the following exponential equation for x:
23x−1= 8
Solution
Step 1: Rewrite 8 as a power of 2 (since the base of the exponential function is
2).
23x−1= 23
Step 2: Set the exponents equal to each other since the bases are the same.
3x−1 = 3
Step 3: Solve for xby isolating it.
3x= 4
Step 4: Divide by 3 to solve for x.
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
Question 27
Question
Solve for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Since 8 = 23, we have:
23x−1= 23
Step 2: Since the bases are the same, we can equate the exponents:
3x−1 = 3
Step 3: Solve for xby isolating x:
3x= 3 + 1
3x= 4
x=4
3
Step 4: Therefore, the solution to the equation 23x−1= 8 is x=4
3.
18
Question 28
Question
Solve the exponential equation algebraically: 53x= 125.
Solution
Step 1: Rewrite 125 as a power of 5. Since 125 = 53, the equation becomes
53x= 53.
Step 2: Set the exponents equal to each other. Since the bases are the same,
we can set the exponents equal to each other: 3x= 3.
Step 3: Solve for x. Dividing both sides by 3, we get
x=3
3= 1.
Step 4: Check the solution. Substitute x= 1 back into the original equation:
53(1) = 53
53= 53.
Both sides are equal, so x= 1 is the solution to the equation.
Question 29
Question
Samantha invests $5000 in a savings account that pays an annual interest rate of
4%, compounded continuously. How long will it take for Samantha’s investment
to double in value?
Solution
Let Arepresent the amount of money in the savings account after tyears.
We know that the formula for continuously compounded interest is given by
A=P ert, where P= $5000 is the principal amount, r= 0.04 is the annual
interest rate, and A= 2P= 2 ×$5000 = $10000.
Step 1: Substitute the given values into the formula:
10000 = 5000e0.04t
Step 2: Divide both sides by 5000:
2 = e0.04t
Step 3: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.04t)
19
Step 4: Use the property ln(ex) = xto simplify the right-hand side:
ln 2 = 0.04t
Step 5: Finally, solve for tby dividing both sides by 0.04:
t=ln 2
0.04 ≈0.6931
0.04 ≈17.33
So, it will take approximately 17.33 years for Samantha’s investment to
double in value when continuously compounded at an annual interest rate of
4%.
Question 30
Question
Let f(x) = 3xand g(x) = log2(x). Find the value of xthat satisfies the equation
f(x) + g(x) = 5.
Solution
Step 1: Substitute the expressions for f(x)and g(x)into the equation f(x) +
g(x) = 5.
3x+ log2(x) = 5
Step 2: Since log2(x)can be rewritten as y= log2(x), we have 2y=x.
Substituting this into the equation, we get:
3x+y= 5
32y+y= 5
Step 3: Since 32y= (32)y= 9y, the equation becomes:
9y+y= 5
Step 4: It is not easy to solve this equation algebraically. We can use a
graphical approach to solve it. Let’s plot the graph of y= 9yand y= 5 and
find their point of intersection.
Step 5: By observing the graphs, we can see that the point of intersection is
approximately y≈0.427.
Step 6: Substituting y≈0.427 back into 2y=x, we get x≈20.427 ≈1.469.
Therefore, the value of xthat satisfies the equation f(x) + g(x) = 5 is
x≈1.469.
20
Question 31
Question
Suppose a population of bacteria triples every 4 hours. If the initial population
is 1000 bacteria, what will be the population after 12 hours?
Solution
Step 1: First, we need to determine the growth factor. Since the population
triples every 4 hours, the growth factor is 3.
Step 2: Next, we can write the exponential growth model using the formula
P(t) = P0·rt, where P(t)is the population at time t,P0is the initial population,
ris the growth factor, and tis the time elapsed.
Step 3: Substituting the given values, we have P(t) = 1000 ·3t/4.
Step 4: To find the population after 12 hours, we substitute t= 12 into the
formula: P(12) = 1000 ·312/4.
Step 5: Simplifying the expression, we get P(12) = 1000 ·33= 1000 ·27 =
27000 bacteria.
Therefore, the population after 12 hours will be 27,000 bacteria.
Question 32
Question
The population of a city is modeled by the function P(t) = 40000·1.025t, where
trepresents the number of years since the year 2020. Calculate the population of
the city in the year 2040 and determine the annual growth rate of the population.
Solution
Step 1: To find the population in the year 2040, we need to substitute t= 20
into the given function P(t):
P(20) = 40000 ·1.02520
Step 2: Calculate the population in the year 2040:
P(20) = 40000 ·1.02520 ≈74218.19
So, the population of the city in the year 2040 is approximately 74218.
Step 3: To determine the annual growth rate of the population, we need to
analyze the base of the exponential function. Since 1.025 is greater than 1, the
population is experiencing exponential growth.
Step 4: The annual growth rate is equal to the base of the exponential
function minus 1, expressed as a percentage. Therefore, the annual growth rate
is:
(1.025 −1) ×100% = 2.5%
21
Step 5: Thus, the annual growth rate of the population is 2.5
Question 33
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the exponent in the equation equal to the exponent in our
rewritten expression.
2x+ 1 = 3
Step 3: Solve for x.
2x= 3 −1
2x= 2
x= 1
Step 4: Check the solution. Substitute x= 1 back into the original equation:
32(1)+1 = 33
33= 27
Since the left side equals the right side, x= 1 is the correct solution.
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 34
Question
Solve the exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the equation equal to the new expression.
32x−1= 33
22
Question 5
Question
Solve for xin the equation 2x−1−3·2x+ 2x+1 = 0.
Solution
Step 1: Rewrite the given equation using properties of exponents:
2x−1−3·2x+ 2x+1 = 0 ⇒2x
2−3·2x
1+ 2 ·2x= 0
Step 2: Find a common denominator to combine the terms:
2x
2−6·2x
2+4·2x
2= 0
Step 3: Combine the terms:
2x−6·2x+ 4 ·2x
2= 0 ⇒−3·2x
2= 0
Step 4: Solve for 2x:
−3·2x= 0 ⇒2x= 0
Since 2xcannot equal 0, there are no solutions to the equation 2x−1−3·
2x+ 2x+1 = 0.
Question 6
Question
Suppose a certain investment grows according to the formula A=P·e0.06t,
where Arepresents the final amount, Pis the principal investment, tis the
time in years, and eis Euler’s number. If an initial investment of $5000 grows
to be $8000, find the time it takes for the investment to double.
Solution
Let’s denote the time it takes for the investment to double as td. To find this
value, we need to set Aequal to 2 times the initial investment (2P) and solve
for td.
Step 1: Set up the equation using the given information. We know that
A= 2P= 2 ·5000 = 10000, so we have:
10000 = 5000 ·e0.06td
4
Step 2: Solve for tdby isolating the variable. Divide both sides by 5000:
e0.06td= 2
Step 3: Take the natural logarithm of both sides to solve for td.
ln(e0.06td)= ln(2)
0.06td= ln(2)
td=ln(2)
0.06
td≈0.693
0.06
td≈11.55
Therefore, it will take approximately 11.55 years for the investment to dou-
ble.
Question 7
Question
The population of a city is growing exponentially at a rate of 3.5% per year. If
the current population is 500,000, what will the population be in 10 years?
Solution
Step 1: Identify the given information and the formula for exponential growth.
Given: - Population growth rate: 3.5% per year - Current population: 500,000
We can model exponential growth using the formula:
P(t) = P0×(1 + r)t
where: - P(t)is the population after tyears, - P0is the initial population, - r
is the growth rate as a decimal, - tis the time in years.
Step 2: Convert the growth rate to a decimal. The growth rate is 3.5%,
which is equivalent to 0.035 as a decimal.
Step 3: Substitute the given values into the formula and solve for the pop-
ulation after 10 years. Plugging in the values:
P(10) = 500,000 ×(1 + 0.035)10
P(10) = 500,000 ×(1.035)10
Step 4: Calculate the population after 10 years.
P(10) = 500,000 ×1.4190675
P(10) ≈709,534
Therefore, the population of the city will be approximately 709,534 in 10
years.
5
Question 8
Question
Solve the following exponential equation for x:3x+1 −2·3x+ 1 = 0.
Solution
Step 1: Let u= 3x. Then the equation becomes u2−2u+ 1 = 0.
Step 2: Solve the quadratic equation u2−2u+ 1 = 0 by factoring:
(u−1)2= 0
Step 3: Set (u−1)2= 0 equal to zero and solve for u:
u−1 = 0 =⇒u= 1
Step 4: Substitute back u= 3xinto u= 1 to get:
3x= 1
Step 5: Solve 3x= 1 to find x:
x= 0
Therefore, the solution to the equation 3x+1 −2·3x+ 1 = 0 is x= 0.
Question 9
Question
Samantha invested $10,000 in a savings account that earns 4% annual interest
compounded quarterly. How long will it take for her investment to double in
value?
Solution
Step 1: The formula for compound interest is given by:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after tyears, including interest,
P= the principal amount (the initial amount of money), r= annual interest
rate (in decimal form), n= number of times that interest is compounded per
year, t= time the money is invested for in years.
Given the information: P= $10,000,r= 0.04,n= 4, Samantha wants to
find twhen A= 2P= 2 ·10000 = $20,000. Substituting these values into the
formula:
20000 = 10000 (1 + 0.04
4)4t
6
Step 2: Simplifying the equation to solve for t:
2 = (1 + 0.01)4t
Step 3: Rewrite the equation:
2 = (1.01)4t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln (1.01)4t
Step 5: Use the property of logarithms to bring down the exponent:
ln(2) = 4t·ln(1.01)
Step 6: Solve for t:
t=ln(2)
4 ln(1.01) ≈0.6931
4·0.0099 ≈17.82 years
Therefore, it will take approximately 17.82 years for Samantha’s investment
to double in value.
Question 10
Question
Samantha invests $5000 in a savings account that earns an annual interest rate of
4% compounded quarterly. Write an exponential function to model the amount
of money A(t)in the account after tyears.
Solution
Step 1: We know that the formula for compound interest is given by:
A(t) = P(1 + r
n)nt
where: - P= $5000 (initial investment) - r= 0.04 (annual interest rate) - n= 4
(number of times interest is compounded in a year) - tis the time in years
Step 2: Substitute the known values into the formula:
A(t) = 5000 (1 + 0.04
4)4t
Step 3: Simplify the expression inside the parentheses:
A(t) = 5000 (1 + 0.01)4t
Step 4: Further simplify the expression:
A(t) = 5000 (1.01)4t
Therefore, the exponential function to model the amount of money A(t)in
the account after tyears is A(t) = 5000(1.01)4t.
7
Question 11
Question
Solve the following exponential equation for x:23x−1−3·2x−1−2 = 0.
Solution
Step 1: Let’s substitute u= 2x−1to simplify the equation. This gives us:
2·u3−3u−2 = 0.
Step 2: Factor the left side of the equation to get: 2u3−4u+u−2=0⇒
2u(u2−2) + 1(u−2) = 0.
Step 3: Set each factor to zero and solve for u. We get u= 2 from u−2 = 0
and u=±√2from u2−2 = 0.
Step 4: Now substitute 2x−1back in for u. So, we have 2x−1= 2 and
2x−1=±√2.
Step 5: Solve the first equation 2x−1= 2 to find x. This simplifies to
x−1 = 1 ⇒x= 2.
Step 6: Solve the second equation 2x−1=±√2to find x. This simplifies to
x−1 = 1
2⇒x=3
2.
Step 7: Therefore, the solutions to the exponential equation 23x−1−3·2x−1−
2 = 0 are x= 2 and x=3
2.
Question 12
Question
Samantha invests $10,000 in an account that earns 4.5% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Step 1: To find the time it takes for an investment to double in value when
interest is compounded continuously, we can use the formula:
A=P·ert
Where: - Ais the amount of money after time t, - Pis the principal amount
(initial investment), - ris the annual interest rate, - tis the time the money is
invested for, and - eis the base of natural logarithms (approximately 2.71828).
Step 2: In this case, Samantha invests $10,000, so P= 10000, the interest
rate is 4.5% or 0.045, and she wants her investment to double, which means
A= 2P= 20000.
Step 3: Substituting the given values into the formula, we have:
20000 = 10000 ·e0.045t
8
Step 4: Divide both sides by 10000 to isolate the exponential expression:
2 = e0.045t
Step 5: To solve for t, we need to take the natural logarithm of both sides:
ln(2) = ln(e0.045t)
Step 6: Since ln(ex) = x, we can simplify our equation to:
ln(2) = 0.045t
Step 7: Now, divide both sides by 0.045 to solve for t:
t=ln(2)
0.045
Step 8: Using a calculator, we find:
t≈0.693147
0.045 ≈15.47
Step 9: Therefore, it will take approximately 15.47 years for Samantha’s
investment to double in value.
Question 13
Question
A certain radioactive substance decays according to the function Q(t) = Q0·
e−0.02t, where Q(t)represents the quantity of the substance remaining after t
years and Q0is the initial quantity of the substance. If the initial quantity of
the substance is 500 grams, find: a) The quantity of the substance that remains
after 10 years. b) The time it takes for the quantity of the substance to drop to
200 grams.
Solution
a) To find the quantity of the substance that remains after 10 years, we can
simply substitute t= 10 into the given exponential decay function.
Q(10) = 500 ·e−0.02·10
Q(10) = 500 ·e−0.2
Q(10) ≈500 ·0.8187
Q(10) ≈409.35 grams
9
b) To find the time it takes for the quantity of the substance to drop to 200
grams, we need to solve the equation:
200 = 500 ·e−0.02t
Dividing both sides by 500:
0.4 = e−0.02t
Taking the natural logarithm of both sides:
ln(0.4) = ln(e−0.02t)
ln(0.4) = −0.02t
Solving for t:
t=ln(0.4)
−0.02
t≈−0.9163
−0.02
t≈45.82 years
Therefore, it takes approximately 45.82 years for the quantity of the sub-
stance to drop to 200 grams.
Question 14
Question
A population of 1000 bacteria doubles in size every 12 hours. Write an expo-
nential growth model to represent the population Pof bacteria as a function of
time tin days. Round your answer to three decimal places.
Solution
Step 1: Since the population of bacteria doubles every 12 hours, the growth rate
is 100% every 12 hours, or 200% in one full day (24 hours). This translates to
a growth factor of 2.
Step 2: We can express the population Pas a function of time tin days
using the formula for exponential growth:
P(t) = P0·2rt,
where: - P0is the initial population, - ris the growth rate (in this case, 1
because the population doubles), - tis the time in days.
Step 3: Since the population starts at 1000 bacteria, we have P0= 1000.
Therefore, the exponential growth model for the population of bacteria is:
P(t) = 1000 ·21t.
10
Step 4: Simplifying the model gives:
P(t) = 1000 ·2t.
Step 5: Rounding the answer to three decimal places, the exponential growth
model for the population of bacteria as a function of time is:
P(t) = 1000 ·2t.
Question 15
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where
tis the time in years. Find the population of the city after 10 years.
Solution
Step 1: To find the population after 10 years, substitute t= 10 into the function
P(t).
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression inside the exponential function.
P(10) = 5000 ·e0.2
Step 3: Calculate the value of e0.2.
P(10) = 5000 ·e0.2= 5000 ·1.221402758
Step 4: Multiply the population by the calculated value of e0.2.
P(10) = 6107.01379
Therefore, the population of the city after 10 years is approximately 6,107
people.
Question 16
Question
A certain radioactive substance decays exponentially. If 100 milligrams of the
substance decays to 70 milligrams in 10 hours, how much of the substance will
remain after 24 hours?
11
Solution
Step 1: Determine the decay constant kusing the given information. Step
2: Use the exponential decay formula to calculate how much of the substance
remains after 24 hours.
Step 1: First, we have the exponential decay formula:
A(t) = A0·e−kt
where: - A(t)is the amount of substance remaining after thours, - A0is the
initial amount of the substance, - kis the decay constant, - eis Euler’s number
(approximately 2.71828).
Given that 100 milligrams decays to 70 milligrams in 10 hours, we can set
up the following equation:
70 = 100 ·e−10k
To solve for k, divide both sides by 100:
0.7 = e−10k
Taking the natural logarithm of both sides:
ln(0.7) = ln(e−10k)
ln(0.7) = −10k
Dividing by -10 to solve for k:
k=−ln(0.7)
10
Step 2: Now that we have found the decay constant k, we can use the
exponential decay formula to calculate how much of the substance remains after
24 hours:
A(24) = 100 ·e−k·24
Substitute k=−ln(0.7)
10 into the formula:
A(24) = 100 ·e−
ln(0.7)
10
·24
Calculating this expression will give us the amount of the substance remain-
ing after 24 hours.
Question 17
Question
Let f(x) = 2x+ 5 and g(x) = log3(x−1). Find the composite function f(g(x)).
12
Solution
Step 1: Find f(g(x)) by substituting g(x)into f(x). Step 2: Rewrite the
composite function in simplified form.
Step 1:
f(g(x)) = f(log3(x−1)) = 2log3(x−1) + 5
Step 2: To simplify 2log3(x−1), we can use the property alogab=b. Hence,
2log3(x−1) =x−1
Therefore, the composite function f(g(x)) simplifies to:
f(g(x)) = x−1 + 5 = x+ 4
So, f(g(x)) = x+ 4.
Question 18
Question
Solve for xin the equation 52x−1= 125.
Solution
Step 1: Rewrite 125 as a power of 5by observing that 125 = 53. Step 2:
Substitute 125 as 53in the equation 52x−1= 125 to get 52x−1= 53. Step 3:
Since the bases are the same, equate the exponents: 2x−1 = 3. Step 4: Solve
the equation 2x−1=3for xby first adding 1to both sides to get 2x= 4.
Step 5: Divide by 2on both sides to solve for x:x= 2. Step 6: Therefore, the
solution to the equation 52x−1= 125 is x= 2.
Question 19
Question
Solve the exponential equation 3x+1 = 27 for x.
Solution
Step 1: Rewrite the base 27 as a power of 3.
27 = 33
Step 2: Substitute 33for 27 in the original equation.
3x+1 = 33
13
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
x+ 1 = 3
Step 4: Solve for xby subtracting 1from both sides.
x= 3 −1
x= 2
Step 5: Therefore, the solution to the exponential equation 3x+1 = 27 is
x= 2.
Question 20
Question
A colony of bacteria doubles in size every 4 hours. If there are initially 100
bacteria in the colony, how many bacteria will there be after 24 hours?
Solution
Step 1: Determine the growth factor for the bacteria colony. Since the colony
doubles in size every 4 hours, the growth factor ris 2.
Step 2: Write the exponential growth model for the bacteria colony. Let
N(t)be the number of bacteria at time tin hours. The model is given by:
N(t) = 100 ×2t
4
Step 3: Substitute t= 24 into the model to find the number of bacteria after
24 hours.
N(24) = 100 ×224
4
Step 4: Simplify the expression.
N(24) = 100 ×26
N(24) = 100 ×64
N(24) = 6400
Step 5: Therefore, after 24 hours, there will be 6400 bacteria in the colony.
Question 21
Question
Solve the exponential equation for x:4x+1 = 32.
14
Solution
Step 1: Rewrite 32 as a power of 2. We have 32 = 25.
Step 2: Rewrite the given exponential equation with a common base, 4 and
2 in this case. 4x+1 = (22)x+1.
Step 3: Apply the power rule of exponents by multiplying the exponents
inside the parentheses. 4x+1 = 22x+2.
Step 4: Since 4x+1 = 22x+2, we can set the exponents equal to each other.
22x+2 = 25.
Step 5: Now equate the exponents. 2x+ 2 = 5.
Step 6: Subtract 2 from both sides of the equation. 2x= 3.
Step 7: Finally, divide by 2 to solve for x.x=3
2.
Therefore, the solution to the exponential equation 4x+1 = 32 is x=3
2.
Question 22
Question
For an investment that earns 5
Solution
Step 1: Let’s start by setting up the equation for when the investment doubles
in value. Since the initial investment is 1000, we want to find the time twhen
the value of the investment is 2000. Therefore, we have:
2000 = 1000(1.05)t
Step 2: Divide both sides by 1000 to isolate the exponential term:
2000
1000 = (1.05)t
Step 3: Simplify the left side of the equation:
2 = (1.05)t
Step 4: To solve for t, we take the natural logarithm (ln) of both sides:
ln(2) = ln((1.05)t)
Step 5: Use the property of logarithms that allows us to bring the exponent
down in front:
ln(2) = tln(1.05)
Step 6: Now, divide both sides by ln(1.05) to solve for t:
t=ln(2)
ln(1.05)
15
Step 7: Use a calculator to find the approximate value of t:
t≈ln(2)
ln(1.05) ≈0.6931
0.0488 ≈14.20
Step 8: Therefore, it will take approximately 14.20 years for the investment
to double in value.
Question 23
Question
The population of a city is currently 50,000 and is projected to double every 10
years. (a) Find the exponential function that models the population of the city
as a function of time in years. (b) Determine the population of the city after 30
years.
Solution
(a) Let P(t)represent the population of the city as a function of time tin
years. Since the population doubles every 10 years, the exponential function
that models the population growth is given by:
P(t) = 50000 ×2t/10
(b) To determine the population of the city after 30 years, we substitute
t= 30 into the exponential function:
P(30) = 50000 ×230/10
P(30) = 50000 ×23
P(30) = 50000 ×8
P(30) = 400000
Therefore, the population of the city after 30 years is 400,000.
Question 24
Question
Solve for x:32x−1= 27.
16
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,the equation becomes 32x−1= 33.
Step 2: Equate the exponents.
Since the bases are equal, we have 2x−1 = 3.
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2.
Step 4: Verify the solution.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
33= 33=⇒27 = 27.
Therefore, the solution is x= 2.
Question 25
Question
Solve the following exponential equation for x:52x−3= 25.
Solution
Step 1: Rewrite 25 as a power of 5.
25 = 52
Step 2: Substitute 52back into the original equation.
52x−3= 52
Step 3: Since the bases are the same, we can set the exponents equal to
each other.
2x−3 = 2
Step 4: Add 3to both sides of the equation.
2x= 5
Step 5: Divide by 2to solve for x.
x=5
2
Therefore, the solution to the equation 52x−3= 25 is x=5
2.
17
Question 26
Question
Solve the following exponential equation for x:
23x−1= 8
Solution
Step 1: Rewrite 8 as a power of 2 (since the base of the exponential function is
2).
23x−1= 23
Step 2: Set the exponents equal to each other since the bases are the same.
3x−1 = 3
Step 3: Solve for xby isolating it.
3x= 4
Step 4: Divide by 3 to solve for x.
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
Question 27
Question
Solve for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Since 8 = 23, we have:
23x−1= 23
Step 2: Since the bases are the same, we can equate the exponents:
3x−1 = 3
Step 3: Solve for xby isolating x:
3x= 3 + 1
3x= 4
x=4
3
Step 4: Therefore, the solution to the equation 23x−1= 8 is x=4
3.
18
Question 28
Question
Solve the exponential equation algebraically: 53x= 125.
Solution
Step 1: Rewrite 125 as a power of 5. Since 125 = 53, the equation becomes
53x= 53.
Step 2: Set the exponents equal to each other. Since the bases are the same,
we can set the exponents equal to each other: 3x= 3.
Step 3: Solve for x. Dividing both sides by 3, we get
x=3
3= 1.
Step 4: Check the solution. Substitute x= 1 back into the original equation:
53(1) = 53
53= 53.
Both sides are equal, so x= 1 is the solution to the equation.
Question 29
Question
Samantha invests $5000 in a savings account that pays an annual interest rate of
4%, compounded continuously. How long will it take for Samantha’s investment
to double in value?
Solution
Let Arepresent the amount of money in the savings account after tyears.
We know that the formula for continuously compounded interest is given by
A=P ert, where P= $5000 is the principal amount, r= 0.04 is the annual
interest rate, and A= 2P= 2 ×$5000 = $10000.
Step 1: Substitute the given values into the formula:
10000 = 5000e0.04t
Step 2: Divide both sides by 5000:
2 = e0.04t
Step 3: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.04t)
19
Step 4: Use the property ln(ex) = xto simplify the right-hand side:
ln 2 = 0.04t
Step 5: Finally, solve for tby dividing both sides by 0.04:
t=ln 2
0.04 ≈0.6931
0.04 ≈17.33
So, it will take approximately 17.33 years for Samantha’s investment to
double in value when continuously compounded at an annual interest rate of
4%.
Question 30
Question
Let f(x) = 3xand g(x) = log2(x). Find the value of xthat satisfies the equation
f(x) + g(x) = 5.
Solution
Step 1: Substitute the expressions for f(x)and g(x)into the equation f(x) +
g(x) = 5.
3x+ log2(x) = 5
Step 2: Since log2(x)can be rewritten as y= log2(x), we have 2y=x.
Substituting this into the equation, we get:
3x+y= 5
32y+y= 5
Step 3: Since 32y= (32)y= 9y, the equation becomes:
9y+y= 5
Step 4: It is not easy to solve this equation algebraically. We can use a
graphical approach to solve it. Let’s plot the graph of y= 9yand y= 5 and
find their point of intersection.
Step 5: By observing the graphs, we can see that the point of intersection is
approximately y≈0.427.
Step 6: Substituting y≈0.427 back into 2y=x, we get x≈20.427 ≈1.469.
Therefore, the value of xthat satisfies the equation f(x) + g(x) = 5 is
x≈1.469.
20
Question 31
Question
Suppose a population of bacteria triples every 4 hours. If the initial population
is 1000 bacteria, what will be the population after 12 hours?
Solution
Step 1: First, we need to determine the growth factor. Since the population
triples every 4 hours, the growth factor is 3.
Step 2: Next, we can write the exponential growth model using the formula
P(t) = P0·rt, where P(t)is the population at time t,P0is the initial population,
ris the growth factor, and tis the time elapsed.
Step 3: Substituting the given values, we have P(t) = 1000 ·3t/4.
Step 4: To find the population after 12 hours, we substitute t= 12 into the
formula: P(12) = 1000 ·312/4.
Step 5: Simplifying the expression, we get P(12) = 1000 ·33= 1000 ·27 =
27000 bacteria.
Therefore, the population after 12 hours will be 27,000 bacteria.
Question 32
Question
The population of a city is modeled by the function P(t) = 40000·1.025t, where
trepresents the number of years since the year 2020. Calculate the population of
the city in the year 2040 and determine the annual growth rate of the population.
Solution
Step 1: To find the population in the year 2040, we need to substitute t= 20
into the given function P(t):
P(20) = 40000 ·1.02520
Step 2: Calculate the population in the year 2040:
P(20) = 40000 ·1.02520 ≈74218.19
So, the population of the city in the year 2040 is approximately 74218.
Step 3: To determine the annual growth rate of the population, we need to
analyze the base of the exponential function. Since 1.025 is greater than 1, the
population is experiencing exponential growth.
Step 4: The annual growth rate is equal to the base of the exponential
function minus 1, expressed as a percentage. Therefore, the annual growth rate
is:
(1.025 −1) ×100% = 2.5%
21
Step 5: Thus, the annual growth rate of the population is 2.5
Question 33
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the exponent in the equation equal to the exponent in our
rewritten expression.
2x+ 1 = 3
Step 3: Solve for x.
2x= 3 −1
2x= 2
x= 1
Step 4: Check the solution. Substitute x= 1 back into the original equation:
32(1)+1 = 33
33= 27
Since the left side equals the right side, x= 1 is the correct solution.
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 34
Question
Solve the exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the equation equal to the new expression.
32x−1= 33
22
Question 5
Question
Solve for xin the equation 2x−1−3·2x+ 2x+1 = 0.
Solution
Step 1: Rewrite the given equation using properties of exponents:
2x−1−3·2x+ 2x+1 = 0 ⇒2x
2−3·2x
1+ 2 ·2x= 0
Step 2: Find a common denominator to combine the terms:
2x
2−6·2x
2+4·2x
2= 0
Step 3: Combine the terms:
2x−6·2x+ 4 ·2x
2= 0 ⇒−3·2x
2= 0
Step 4: Solve for 2x:
−3·2x= 0 ⇒2x= 0
Since 2xcannot equal 0, there are no solutions to the equation 2x−1−3·
2x+ 2x+1 = 0.
Question 6
Question
Suppose a certain investment grows according to the formula A=P·e0.06t,
where Arepresents the final amount, Pis the principal investment, tis the
time in years, and eis Euler’s number. If an initial investment of $5000 grows
to be $8000, find the time it takes for the investment to double.
Solution
Let’s denote the time it takes for the investment to double as td. To find this
value, we need to set Aequal to 2 times the initial investment (2P) and solve
for td.
Step 1: Set up the equation using the given information. We know that
A= 2P= 2 ·5000 = 10000, so we have:
10000 = 5000 ·e0.06td
4
Step 2: Solve for tdby isolating the variable. Divide both sides by 5000:
e0.06td= 2
Step 3: Take the natural logarithm of both sides to solve for td.
ln(e0.06td)= ln(2)
0.06td= ln(2)
td=ln(2)
0.06
td≈0.693
0.06
td≈11.55
Therefore, it will take approximately 11.55 years for the investment to dou-
ble.
Question 7
Question
The population of a city is growing exponentially at a rate of 3.5% per year. If
the current population is 500,000, what will the population be in 10 years?
Solution
Step 1: Identify the given information and the formula for exponential growth.
Given: - Population growth rate: 3.5% per year - Current population: 500,000
We can model exponential growth using the formula:
P(t) = P0×(1 + r)t
where: - P(t)is the population after tyears, - P0is the initial population, - r
is the growth rate as a decimal, - tis the time in years.
Step 2: Convert the growth rate to a decimal. The growth rate is 3.5%,
which is equivalent to 0.035 as a decimal.
Step 3: Substitute the given values into the formula and solve for the pop-
ulation after 10 years. Plugging in the values:
P(10) = 500,000 ×(1 + 0.035)10
P(10) = 500,000 ×(1.035)10
Step 4: Calculate the population after 10 years.
P(10) = 500,000 ×1.4190675
P(10) ≈709,534
Therefore, the population of the city will be approximately 709,534 in 10
years.
5
Question 8
Question
Solve the following exponential equation for x:3x+1 −2·3x+ 1 = 0.
Solution
Step 1: Let u= 3x. Then the equation becomes u2−2u+ 1 = 0.
Step 2: Solve the quadratic equation u2−2u+ 1 = 0 by factoring:
(u−1)2= 0
Step 3: Set (u−1)2= 0 equal to zero and solve for u:
u−1 = 0 =⇒u= 1
Step 4: Substitute back u= 3xinto u= 1 to get:
3x= 1
Step 5: Solve 3x= 1 to find x:
x= 0
Therefore, the solution to the equation 3x+1 −2·3x+ 1 = 0 is x= 0.
Question 9
Question
Samantha invested $10,000 in a savings account that earns 4% annual interest
compounded quarterly. How long will it take for her investment to double in
value?
Solution
Step 1: The formula for compound interest is given by:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after tyears, including interest,
P= the principal amount (the initial amount of money), r= annual interest
rate (in decimal form), n= number of times that interest is compounded per
year, t= time the money is invested for in years.
Given the information: P= $10,000,r= 0.04,n= 4, Samantha wants to
find twhen A= 2P= 2 ·10000 = $20,000. Substituting these values into the
formula:
20000 = 10000 (1 + 0.04
4)4t
6
Step 2: Simplifying the equation to solve for t:
2 = (1 + 0.01)4t
Step 3: Rewrite the equation:
2 = (1.01)4t
Step 4: Take the natural logarithm of both sides to solve for t:
ln(2) = ln (1.01)4t
Step 5: Use the property of logarithms to bring down the exponent:
ln(2) = 4t·ln(1.01)
Step 6: Solve for t:
t=ln(2)
4 ln(1.01) ≈0.6931
4·0.0099 ≈17.82 years
Therefore, it will take approximately 17.82 years for Samantha’s investment
to double in value.
Question 10
Question
Samantha invests $5000 in a savings account that earns an annual interest rate of
4% compounded quarterly. Write an exponential function to model the amount
of money A(t)in the account after tyears.
Solution
Step 1: We know that the formula for compound interest is given by:
A(t) = P(1 + r
n)nt
where: - P= $5000 (initial investment) - r= 0.04 (annual interest rate) - n= 4
(number of times interest is compounded in a year) - tis the time in years
Step 2: Substitute the known values into the formula:
A(t) = 5000 (1 + 0.04
4)4t
Step 3: Simplify the expression inside the parentheses:
A(t) = 5000 (1 + 0.01)4t
Step 4: Further simplify the expression:
A(t) = 5000 (1.01)4t
Therefore, the exponential function to model the amount of money A(t)in
the account after tyears is A(t) = 5000(1.01)4t.
7
Question 11
Question
Solve the following exponential equation for x:23x−1−3·2x−1−2 = 0.
Solution
Step 1: Let’s substitute u= 2x−1to simplify the equation. This gives us:
2·u3−3u−2 = 0.
Step 2: Factor the left side of the equation to get: 2u3−4u+u−2=0⇒
2u(u2−2) + 1(u−2) = 0.
Step 3: Set each factor to zero and solve for u. We get u= 2 from u−2 = 0
and u=±√2from u2−2 = 0.
Step 4: Now substitute 2x−1back in for u. So, we have 2x−1= 2 and
2x−1=±√2.
Step 5: Solve the first equation 2x−1= 2 to find x. This simplifies to
x−1 = 1 ⇒x= 2.
Step 6: Solve the second equation 2x−1=±√2to find x. This simplifies to
x−1 = 1
2⇒x=3
2.
Step 7: Therefore, the solutions to the exponential equation 23x−1−3·2x−1−
2 = 0 are x= 2 and x=3
2.
Question 12
Question
Samantha invests $10,000 in an account that earns 4.5% interest compounded
continuously. How long will it take for her investment to double in value?
Solution
Step 1: To find the time it takes for an investment to double in value when
interest is compounded continuously, we can use the formula:
A=P·ert
Where: - Ais the amount of money after time t, - Pis the principal amount
(initial investment), - ris the annual interest rate, - tis the time the money is
invested for, and - eis the base of natural logarithms (approximately 2.71828).
Step 2: In this case, Samantha invests $10,000, so P= 10000, the interest
rate is 4.5% or 0.045, and she wants her investment to double, which means
A= 2P= 20000.
Step 3: Substituting the given values into the formula, we have:
20000 = 10000 ·e0.045t
8
Step 4: Divide both sides by 10000 to isolate the exponential expression:
2 = e0.045t
Step 5: To solve for t, we need to take the natural logarithm of both sides:
ln(2) = ln(e0.045t)
Step 6: Since ln(ex) = x, we can simplify our equation to:
ln(2) = 0.045t
Step 7: Now, divide both sides by 0.045 to solve for t:
t=ln(2)
0.045
Step 8: Using a calculator, we find:
t≈0.693147
0.045 ≈15.47
Step 9: Therefore, it will take approximately 15.47 years for Samantha’s
investment to double in value.
Question 13
Question
A certain radioactive substance decays according to the function Q(t) = Q0·
e−0.02t, where Q(t)represents the quantity of the substance remaining after t
years and Q0is the initial quantity of the substance. If the initial quantity of
the substance is 500 grams, find: a) The quantity of the substance that remains
after 10 years. b) The time it takes for the quantity of the substance to drop to
200 grams.
Solution
a) To find the quantity of the substance that remains after 10 years, we can
simply substitute t= 10 into the given exponential decay function.
Q(10) = 500 ·e−0.02·10
Q(10) = 500 ·e−0.2
Q(10) ≈500 ·0.8187
Q(10) ≈409.35 grams
9
b) To find the time it takes for the quantity of the substance to drop to 200
grams, we need to solve the equation:
200 = 500 ·e−0.02t
Dividing both sides by 500:
0.4 = e−0.02t
Taking the natural logarithm of both sides:
ln(0.4) = ln(e−0.02t)
ln(0.4) = −0.02t
Solving for t:
t=ln(0.4)
−0.02
t≈−0.9163
−0.02
t≈45.82 years
Therefore, it takes approximately 45.82 years for the quantity of the sub-
stance to drop to 200 grams.
Question 14
Question
A population of 1000 bacteria doubles in size every 12 hours. Write an expo-
nential growth model to represent the population Pof bacteria as a function of
time tin days. Round your answer to three decimal places.
Solution
Step 1: Since the population of bacteria doubles every 12 hours, the growth rate
is 100% every 12 hours, or 200% in one full day (24 hours). This translates to
a growth factor of 2.
Step 2: We can express the population Pas a function of time tin days
using the formula for exponential growth:
P(t) = P0·2rt,
where: - P0is the initial population, - ris the growth rate (in this case, 1
because the population doubles), - tis the time in days.
Step 3: Since the population starts at 1000 bacteria, we have P0= 1000.
Therefore, the exponential growth model for the population of bacteria is:
P(t) = 1000 ·21t.
10
Step 4: Simplifying the model gives:
P(t) = 1000 ·2t.
Step 5: Rounding the answer to three decimal places, the exponential growth
model for the population of bacteria as a function of time is:
P(t) = 1000 ·2t.
Question 15
Question
The population of a city is modeled by the function P(t) = 5000 ·e0.02t, where
tis the time in years. Find the population of the city after 10 years.
Solution
Step 1: To find the population after 10 years, substitute t= 10 into the function
P(t).
P(10) = 5000 ·e0.02·10
Step 2: Simplify the expression inside the exponential function.
P(10) = 5000 ·e0.2
Step 3: Calculate the value of e0.2.
P(10) = 5000 ·e0.2= 5000 ·1.221402758
Step 4: Multiply the population by the calculated value of e0.2.
P(10) = 6107.01379
Therefore, the population of the city after 10 years is approximately 6,107
people.
Question 16
Question
A certain radioactive substance decays exponentially. If 100 milligrams of the
substance decays to 70 milligrams in 10 hours, how much of the substance will
remain after 24 hours?
11
Solution
Step 1: Determine the decay constant kusing the given information. Step
2: Use the exponential decay formula to calculate how much of the substance
remains after 24 hours.
Step 1: First, we have the exponential decay formula:
A(t) = A0·e−kt
where: - A(t)is the amount of substance remaining after thours, - A0is the
initial amount of the substance, - kis the decay constant, - eis Euler’s number
(approximately 2.71828).
Given that 100 milligrams decays to 70 milligrams in 10 hours, we can set
up the following equation:
70 = 100 ·e−10k
To solve for k, divide both sides by 100:
0.7 = e−10k
Taking the natural logarithm of both sides:
ln(0.7) = ln(e−10k)
ln(0.7) = −10k
Dividing by -10 to solve for k:
k=−ln(0.7)
10
Step 2: Now that we have found the decay constant k, we can use the
exponential decay formula to calculate how much of the substance remains after
24 hours:
A(24) = 100 ·e−k·24
Substitute k=−ln(0.7)
10 into the formula:
A(24) = 100 ·e−
ln(0.7)
10
·24
Calculating this expression will give us the amount of the substance remain-
ing after 24 hours.
Question 17
Question
Let f(x) = 2x+ 5 and g(x) = log3(x−1). Find the composite function f(g(x)).
12
Solution
Step 1: Find f(g(x)) by substituting g(x)into f(x). Step 2: Rewrite the
composite function in simplified form.
Step 1:
f(g(x)) = f(log3(x−1)) = 2log3(x−1) + 5
Step 2: To simplify 2log3(x−1), we can use the property alogab=b. Hence,
2log3(x−1) =x−1
Therefore, the composite function f(g(x)) simplifies to:
f(g(x)) = x−1 + 5 = x+ 4
So, f(g(x)) = x+ 4.
Question 18
Question
Solve for xin the equation 52x−1= 125.
Solution
Step 1: Rewrite 125 as a power of 5by observing that 125 = 53. Step 2:
Substitute 125 as 53in the equation 52x−1= 125 to get 52x−1= 53. Step 3:
Since the bases are the same, equate the exponents: 2x−1 = 3. Step 4: Solve
the equation 2x−1=3for xby first adding 1to both sides to get 2x= 4.
Step 5: Divide by 2on both sides to solve for x:x= 2. Step 6: Therefore, the
solution to the equation 52x−1= 125 is x= 2.
Question 19
Question
Solve the exponential equation 3x+1 = 27 for x.
Solution
Step 1: Rewrite the base 27 as a power of 3.
27 = 33
Step 2: Substitute 33for 27 in the original equation.
3x+1 = 33
13
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
x+ 1 = 3
Step 4: Solve for xby subtracting 1from both sides.
x= 3 −1
x= 2
Step 5: Therefore, the solution to the exponential equation 3x+1 = 27 is
x= 2.
Question 20
Question
A colony of bacteria doubles in size every 4 hours. If there are initially 100
bacteria in the colony, how many bacteria will there be after 24 hours?
Solution
Step 1: Determine the growth factor for the bacteria colony. Since the colony
doubles in size every 4 hours, the growth factor ris 2.
Step 2: Write the exponential growth model for the bacteria colony. Let
N(t)be the number of bacteria at time tin hours. The model is given by:
N(t) = 100 ×2t
4
Step 3: Substitute t= 24 into the model to find the number of bacteria after
24 hours.
N(24) = 100 ×224
4
Step 4: Simplify the expression.
N(24) = 100 ×26
N(24) = 100 ×64
N(24) = 6400
Step 5: Therefore, after 24 hours, there will be 6400 bacteria in the colony.
Question 21
Question
Solve the exponential equation for x:4x+1 = 32.
14
Solution
Step 1: Rewrite 32 as a power of 2. We have 32 = 25.
Step 2: Rewrite the given exponential equation with a common base, 4 and
2 in this case. 4x+1 = (22)x+1.
Step 3: Apply the power rule of exponents by multiplying the exponents
inside the parentheses. 4x+1 = 22x+2.
Step 4: Since 4x+1 = 22x+2, we can set the exponents equal to each other.
22x+2 = 25.
Step 5: Now equate the exponents. 2x+ 2 = 5.
Step 6: Subtract 2 from both sides of the equation. 2x= 3.
Step 7: Finally, divide by 2 to solve for x.x=3
2.
Therefore, the solution to the exponential equation 4x+1 = 32 is x=3
2.
Question 22
Question
For an investment that earns 5
Solution
Step 1: Let’s start by setting up the equation for when the investment doubles
in value. Since the initial investment is 1000, we want to find the time twhen
the value of the investment is 2000. Therefore, we have:
2000 = 1000(1.05)t
Step 2: Divide both sides by 1000 to isolate the exponential term:
2000
1000 = (1.05)t
Step 3: Simplify the left side of the equation:
2 = (1.05)t
Step 4: To solve for t, we take the natural logarithm (ln) of both sides:
ln(2) = ln((1.05)t)
Step 5: Use the property of logarithms that allows us to bring the exponent
down in front:
ln(2) = tln(1.05)
Step 6: Now, divide both sides by ln(1.05) to solve for t:
t=ln(2)
ln(1.05)
15
Step 7: Use a calculator to find the approximate value of t:
t≈ln(2)
ln(1.05) ≈0.6931
0.0488 ≈14.20
Step 8: Therefore, it will take approximately 14.20 years for the investment
to double in value.
Question 23
Question
The population of a city is currently 50,000 and is projected to double every 10
years. (a) Find the exponential function that models the population of the city
as a function of time in years. (b) Determine the population of the city after 30
years.
Solution
(a) Let P(t)represent the population of the city as a function of time tin
years. Since the population doubles every 10 years, the exponential function
that models the population growth is given by:
P(t) = 50000 ×2t/10
(b) To determine the population of the city after 30 years, we substitute
t= 30 into the exponential function:
P(30) = 50000 ×230/10
P(30) = 50000 ×23
P(30) = 50000 ×8
P(30) = 400000
Therefore, the population of the city after 30 years is 400,000.
Question 24
Question
Solve for x:32x−1= 27.
16
Solution
Step 1: Rewrite 27 as a power of 3.
Since 27 = 33,the equation becomes 32x−1= 33.
Step 2: Equate the exponents.
Since the bases are equal, we have 2x−1 = 3.
Step 3: Solve for x.
2x−1 = 3
2x= 4
x= 2.
Step 4: Verify the solution.
Substitute x= 2 back into the original equation: 32(2)−1= 33.
33= 33=⇒27 = 27.
Therefore, the solution is x= 2.
Question 25
Question
Solve the following exponential equation for x:52x−3= 25.
Solution
Step 1: Rewrite 25 as a power of 5.
25 = 52
Step 2: Substitute 52back into the original equation.
52x−3= 52
Step 3: Since the bases are the same, we can set the exponents equal to
each other.
2x−3 = 2
Step 4: Add 3to both sides of the equation.
2x= 5
Step 5: Divide by 2to solve for x.
x=5
2
Therefore, the solution to the equation 52x−3= 25 is x=5
2.
17
Question 26
Question
Solve the following exponential equation for x:
23x−1= 8
Solution
Step 1: Rewrite 8 as a power of 2 (since the base of the exponential function is
2).
23x−1= 23
Step 2: Set the exponents equal to each other since the bases are the same.
3x−1 = 3
Step 3: Solve for xby isolating it.
3x= 4
Step 4: Divide by 3 to solve for x.
x=4
3
Therefore, the solution to the equation 23x−1= 8 is x=4
3.
Question 27
Question
Solve for x:23x−1= 8.
Solution
Step 1: Rewrite 8as a power of 2. Since 8 = 23, we have:
23x−1= 23
Step 2: Since the bases are the same, we can equate the exponents:
3x−1 = 3
Step 3: Solve for xby isolating x:
3x= 3 + 1
3x= 4
x=4
3
Step 4: Therefore, the solution to the equation 23x−1= 8 is x=4
3.
18
Question 28
Question
Solve the exponential equation algebraically: 53x= 125.
Solution
Step 1: Rewrite 125 as a power of 5. Since 125 = 53, the equation becomes
53x= 53.
Step 2: Set the exponents equal to each other. Since the bases are the same,
we can set the exponents equal to each other: 3x= 3.
Step 3: Solve for x. Dividing both sides by 3, we get
x=3
3= 1.
Step 4: Check the solution. Substitute x= 1 back into the original equation:
53(1) = 53
53= 53.
Both sides are equal, so x= 1 is the solution to the equation.
Question 29
Question
Samantha invests $5000 in a savings account that pays an annual interest rate of
4%, compounded continuously. How long will it take for Samantha’s investment
to double in value?
Solution
Let Arepresent the amount of money in the savings account after tyears.
We know that the formula for continuously compounded interest is given by
A=P ert, where P= $5000 is the principal amount, r= 0.04 is the annual
interest rate, and A= 2P= 2 ×$5000 = $10000.
Step 1: Substitute the given values into the formula:
10000 = 5000e0.04t
Step 2: Divide both sides by 5000:
2 = e0.04t
Step 3: Take the natural logarithm of both sides to solve for t:
ln 2 = ln(e0.04t)
19
Step 4: Use the property ln(ex) = xto simplify the right-hand side:
ln 2 = 0.04t
Step 5: Finally, solve for tby dividing both sides by 0.04:
t=ln 2
0.04 ≈0.6931
0.04 ≈17.33
So, it will take approximately 17.33 years for Samantha’s investment to
double in value when continuously compounded at an annual interest rate of
4%.
Question 30
Question
Let f(x) = 3xand g(x) = log2(x). Find the value of xthat satisfies the equation
f(x) + g(x) = 5.
Solution
Step 1: Substitute the expressions for f(x)and g(x)into the equation f(x) +
g(x) = 5.
3x+ log2(x) = 5
Step 2: Since log2(x)can be rewritten as y= log2(x), we have 2y=x.
Substituting this into the equation, we get:
3x+y= 5
32y+y= 5
Step 3: Since 32y= (32)y= 9y, the equation becomes:
9y+y= 5
Step 4: It is not easy to solve this equation algebraically. We can use a
graphical approach to solve it. Let’s plot the graph of y= 9yand y= 5 and
find their point of intersection.
Step 5: By observing the graphs, we can see that the point of intersection is
approximately y≈0.427.
Step 6: Substituting y≈0.427 back into 2y=x, we get x≈20.427 ≈1.469.
Therefore, the value of xthat satisfies the equation f(x) + g(x) = 5 is
x≈1.469.
20
Question 31
Question
Suppose a population of bacteria triples every 4 hours. If the initial population
is 1000 bacteria, what will be the population after 12 hours?
Solution
Step 1: First, we need to determine the growth factor. Since the population
triples every 4 hours, the growth factor is 3.
Step 2: Next, we can write the exponential growth model using the formula
P(t) = P0·rt, where P(t)is the population at time t,P0is the initial population,
ris the growth factor, and tis the time elapsed.
Step 3: Substituting the given values, we have P(t) = 1000 ·3t/4.
Step 4: To find the population after 12 hours, we substitute t= 12 into the
formula: P(12) = 1000 ·312/4.
Step 5: Simplifying the expression, we get P(12) = 1000 ·33= 1000 ·27 =
27000 bacteria.
Therefore, the population after 12 hours will be 27,000 bacteria.
Question 32
Question
The population of a city is modeled by the function P(t) = 40000·1.025t, where
trepresents the number of years since the year 2020. Calculate the population of
the city in the year 2040 and determine the annual growth rate of the population.
Solution
Step 1: To find the population in the year 2040, we need to substitute t= 20
into the given function P(t):
P(20) = 40000 ·1.02520
Step 2: Calculate the population in the year 2040:
P(20) = 40000 ·1.02520 ≈74218.19
So, the population of the city in the year 2040 is approximately 74218.
Step 3: To determine the annual growth rate of the population, we need to
analyze the base of the exponential function. Since 1.025 is greater than 1, the
population is experiencing exponential growth.
Step 4: The annual growth rate is equal to the base of the exponential
function minus 1, expressed as a percentage. Therefore, the annual growth rate
is:
(1.025 −1) ×100% = 2.5%
21
Step 5: Thus, the annual growth rate of the population is 2.5
Question 33
Question
Solve the exponential equation 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the exponent in the equation equal to the exponent in our
rewritten expression.
2x+ 1 = 3
Step 3: Solve for x.
2x= 3 −1
2x= 2
x= 1
Step 4: Check the solution. Substitute x= 1 back into the original equation:
32(1)+1 = 33
33= 27
Since the left side equals the right side, x= 1 is the correct solution.
Therefore, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 34
Question
Solve the exponential equation for x:32x−1= 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Set the equation equal to the new expression.
32x−1= 33
22
Step 3: Since the bases are the same, the exponents must be equal.
2x−1 = 3
Step 4: Solve for x.
2x= 4
x= 2
Therefore, the solution to the equation 32x−1= 27 is x= 2.
Question 35
Question
Solve the equation 5x+2 −5x= 50.
Solution
Step 1: Rewrite the equation using the properties of exponents: 5x·52−5x= 50.
Step 2: Simplify the left side: 25 ·5x−5x= 50.
Step 3: Combine like terms on the left side: 24 ·5x= 50.
Step 4: Divide both sides by 24 to isolate 5x:5x=50
24 .
Step 5: Simplify the right side: 5x=25
12 .
Step 6: Take the logarithm of both sides to solve for x:log5(5x) = log5(25
12 ).
Step 7: Use the property of logarithms to bring down the exponent: xlog5(5) =
log5(25
12 ).
Step 8: Simplify the left side: x= log5(25
12 ).
Step 9: Use the change of base formula to rewrite the logarithm: x=ln(25
12 )
ln(5) .
Step 10: Use a calculator to find the approximate value of x:x≈ −0.267.
Therefore, the solution to the equation 5x+2 −5x= 50 is x≈ −0.267.
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