MATH 121 - COLLEGE ALGEBRA -
Applications of exponential and
logarithmic functions
Question Bank - Set 3
Liberty University
Question 1
Question
Solve the following exponential equation for x:3x−1−2·3x+ 3 = 0.
Solution
Step 1: Let’s rewrite the equation in a more familiar form by noticing that 3x−1
can be rewritten as 1
3·3x. This gives us:
1
3·3x−2·3x+ 3 = 0
Step 2: To simplify the equation further, let’s multiply through by 3to clear
the fractions:
1·3x−2·3x+1 + 3 ·3 = 0
Step 3: Simplify the equation to get:
3x−6·3x+ 9 = 0
Step 4: Combine like terms to obtain:
−5·3x+ 9 = 0
Step 5: Add 5·3xto both sides:
9 = 5 ·3x
Step 6: Divide by 5to solve for 3x:
3x=9
5
Step 7: Rewrite 9
5as 3log3
9
5:
3x= 3log3
9
5
Step 8: Since the bases are the same, we can set the exponents equal to each
other:
x= log3
9
5
Therefore, the solution to the exponential equation 3x−1−2·3x+ 3 = 0 is
x= log39
5.
Question 2
Question
Solve the equation 3x−1= 27 for x.
Solution
Step 1: Write 27 as a power of 3:
27 = 33
Step 2: Substitute 27 as 33in the equation and solve for x:
3x−1= 33
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 adding 1to both sides:
x= 4
Step 5: Thus, the solution to the equation 3x−1= 27 is x= 4.
Question 3
Question
Samantha invests $5000 in a savings account that pays an annual interest rate
of 4.5%, compounded continuously. How long will it take for her investment to
double?
2
Solution
Step 1: We can model Samantha’s investment using the continuous compound
interest formula:
A=P·ert
where: - Ais the amount after time t, - Pis the principal amount (initial
investment), - ris the annual interest rate, - tis the time in years, and - eis
the base of the natural logarithm.
Step 2: Since Samantha wants to double her investment, the amount after
doubling will be $10,000. Thus, we have:
10000 = 5000 ·e0.045t
Step 3: Divide both sides by 5000 to solve for the exponential equation:
2 = e0.045t
Step 4: To solve for t, we take the natural logarithm of both sides of the
equation:
ln(2) = ln(e0.045t)
Step 5: Use the property of logarithms to bring down the exponent:
ln(2) = 0.045tln(e)
Step 6: Recall that ln(e) = 1, so our equation simplifies to:
ln(2) = 0.045t
Step 7: Now, solve for tby dividing both sides by 0.045:
t=ln(2)
0.045
Step 8: Use a calculator to approximate the value of t:
t≈ln(2)
0.045 ≈0.6931
0.045 ≈15.47
Step 9: Therefore, it will take approximately 15.47 years for Samantha’s
investment to double.
Question 4
Question
Solve the exponential equation: 3x−27 = 0.
3
Solution
Step 1: Add 27 to both sides of the equation to isolate the exponential term:
3x= 27
Step 2: Rewrite 27 as 33:
3x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
x= 3
Step 4: Therefore, the solution to the exponential equation 3x−27 = 0 is
x= 3.
Question 5
Question
Samantha invested $10,000 in a savings account that pays an annual interest
rate of 4.5%, compounded continuously. How long will it take for her investment
to double in value?
Solution
Let Pbe the initial investment of $10,000, rbe the annual interest rate (4.5%
or 0.045), and tbe the time in years it takes for the investment to double.
We can use the continuous compound interest formula to model the situation:
A=P ert
Since Samantha’s investment needs to double, the final amount Awill be 2P:
2P=P e0.045t
Dividing both sides by P, we get:
2 = e0.045t
To solve for t, we take the natural logarithm of both sides:
ln 2 = ln e0.045t
ln 2 = 0.045t
t=ln 2
0.045
Step 1: Calculate ln 2.
ln 2 ≈0.693
4
Step 2: Plug ln 2 into the formula to find t.
t=0.693
0.045
t≈15.4years
Therefore, it will take approximately 15.4 years for Samantha’s investment
to double in value when compounded continuously.
Question 6
Question
Samantha invested $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much will Samantha have in the account after 10
years?
Solution
Step 1: To find the amount of money in the account after 10 years, we can use
the continuous compounding formula:
A=P·ert
where: - Ais the amount of money in the account after tyears, - Pis the
principal amount initially invested ($5000 in this case), - ris the annual interest
rate (3.5% or 0.035 as a decimal), - tis the time the money is invested for (10
years), and - eis the base of the natural logarithm (approximately 2.71828).
Step 2: Substituting the given values into the formula, we get:
A= 5000 ·e0.035·10
Step 3: Calculate the exponential term:
A≈5000 ·e0.35
Step 4: Evaluate the exponential term:
A≈5000 ·1.4203
Step 5: Multiply to find the final amount:
A≈$7101.50
Therefore, Samantha will have approximately $7101.50 in the account after
10 years.
5
Question 7
Question
Solve for x:5x−1= 125
Solution
Step 1: Rewrite 125 as a power of 5.
125 = 53
Step 2: Substitute 125 with 53in the equation.
5x−1= 53
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
x−1 = 3
Step 4: Add 1 to both sides to solve for x.
x= 4
Thus, the solution to the equation is x= 4.
Question 8
Question
Solve the exponential equation: 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 = 33into the equation 32x+1 = 27.
32x+1 = 33
Step 3: Apply the property of equality for exponents, am=animplies
m=n, to solve for x+ 1.
2x+ 1 = 3
6
Step 4: Solve for xby isolating x.
2x= 3 −1
2x= 2
x= 1
Step 5: Finally, verify the solution x= 1 by substituting it back into the
original equation.
32(1)+1 = 27
33= 27
27 = 27
Thus, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 9
Question
Solve the exponential equation 2x+1 = 8.
Solution
Step 1: Rewrite 8 as a power of 2.
Since 8 = 23,we have 2x+1 = 23.
Step 2: Set the exponents equal to each other.
Therefore, x+ 1 = 3.
Step 3: Solve for x.
Subtracting 1 from both sides gives us x= 2.
Step 4: Verify the solution.
Substitute x= 2 back into the original equation: 22+1 = 23= 8,
which confirms that x= 2 is the correct solution.
Question 10
Question
Samantha invested $5000 in an account earning 3.5% annual interest com-
pounded continuously. How much will she have in the account after 10 years?
7
Solution
Step 1: The formula for compound interest compounded continuously is given
by A=P·ert, where: - Ais the amount of money accumulated after tyears,
-Pis the principal amount (the initial amount of money), - ris the annual
interest rate (written as a decimal), - tis the time the money is invested for (in
years), and - eis the base of the natural logarithm.
Step 2: Given that the principal amount P= $5000, the interest rate r=
0.035 (3.5% written as a decimal), and the time t= 10 years, we can plug these
values into the formula:
A= 5000 ·e0.035·10
Step 3: Simplifying the formula:
A= 5000 ·e0.35
Step 4: Calculating e0.35:
A≈5000 ·1.419067
Step 5: Finding the final amount in the account after 10 years:
A≈7095.335
Therefore, after 10 years, Samantha will have approximately $7095.34 in the
account.
Question 11
Question
Suzanne invests $200 in a savings account that pays 4% interest compounded
quarterly. How much money will be in the account after 5 years?
Solution
Step 1: First, we need to determine the annual interest rate (r) and the number
of compounding periods per year (n). The formula for compound interest is:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= annual interest rate (as a decimal), n= number
of compounding periods per year, and t= time in years. Given that the interest
is compounded quarterly, n= 4 and r= 0.04.
Step 2: Let’s substitute the values into the formula and calculate the amount
of money in the account after 5 years:
A= 200 (1 + 0.04
4)4·5
8
A= 200 (1 + 0.01)20
A= 200 ×1.0120
Step 3: Now, we compute 1.0120 to find the final amount in the account
after 5 years.
A≈200 ×1.221386
A≈244.2772
Therefore, after 5 years, Suzanne will have approximately $244.28 in the
savings account.
Question 12
Question
A certain amount of money is invested at an annual interest rate of 5
Solution
Step 1: Given that the initial investment doubles in 10 years, we can express
this information using the continuous compound interest formula:
A=P·ert
where: - Ais the final amount after time t, - Pis the initial amount invested,
-ris the annual interest rate (decimal form), and - tis the time the money is
invested for.
Step 2: Since the initial investment doubles, the final amount after 10 years
is twice the initial amount:
2P=P·e0.05·10
Step 3: Simplify the equation from Step 2 to solve for P:
2 = e0.5
Step 4: Take natural logarithm (ln) of both sides of the equation to eliminate
the exponential:
ln(2) = ln(e0.5)
Step 5: Use the property ln(ex) = xto simplify the equation from Step 4:
ln(2) = 0.5
Step 6: Solve for Pby converting the equation in Step 5 back to exponential
form:
e0.5= 2
Step 7: Therefore, the initial amount of money invested was 2.
9
Question 13
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4% compounded continuously. How much will her investment be worth after
10 years?
Solution
Step 1: Identify the variables and known values in the problem. Let’s denote
the initial amount invested as P= $5000, the annual interest rate as r= 0.04,
the number of years as t= 10, and the amount after tyears as A.
Step 2: Use the formula for continuous compounding to find the future value
of the investment. The formula for continuous compounding is given by:
A=P·ert
Step 3: Substitute the values into the formula and solve for A. Substitute
P= 5000,r= 0.04, and t= 10 into the formula:
A= 5000 ·e0.04·10
Step 4: Calculate the value of A.
A= 5000 ·e0.4
A= 5000 ·e0.4≈5000 ·1.4918 ≈7459.08
Therefore, after 10 years, Samantha’s investment will be worth approxi-
mately $7459.08.
Question 14
Question
Samantha invested some money in a savings account that earns an annual inter-
est rate of 6.5%, compounded quarterly. If she initially invested $10,000, how
much money will she have in the account after 5 years?
Solution
Step 1: Identify the relevant compound interest formula. The formula for com-
pound interest is given by:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after n years, including interest,
P= the principal amount (the initial amount of money), r= annual interest
10
rate (decimal), n= number of times that interest is compounded per year, t=
time the money is invested for in years.
Step 2: Plug in the known values into the formula. In this case, we have:
P= $10,000 (initial investment), r= 6.5% = 0.065 (annual interest rate in
decimal form), n= 4 (interest is compounded quarterly), t= 5 years.
Step 3: Calculate the amount of money Samantha will have in the account
after 5 years using the compound interest formula:
A= 10000 (1 + 0.065
4)4×5
Step 4: Simplify the expression inside the parentheses first:
1 + 0.065
4= 1 + 0.01625 = 1.01625
Step 5: Plug this back into the formula and calculate:
A= 10000 ×(1.01625)20
Step 6: Evaluate (1.01625)20:
(1.01625)20 ≈1.3548
Step 7: Finally, calculate the total amount of money Samantha will have
after 5 years:
A= 10000 ×1.3548 ≈$13,548
Therefore, after 5 years, Samantha will have approximately $13,548 in her
savings account.
Question 15
Question
Solve the exponential equation 32x−6·3x+ 9 = 0.
Solution
Step 1: Let’s make a substitution to simplify the equation. Let y= 3x. Then,
the original equation becomes y2−6y+ 9 = 0.
Step 2: Now, we need to solve the quadratic equation y2−6y+ 9 = 0.
Step 3: Factoring the quadratic equation gives us (y−3)2= 0.
Step 4: To find the value of y, we take the square root of both sides: y−3 = 0.
Step 5: Therefore, y= 3.
Step 6: Now, substitute back 3xfor y:3x= 3.
Step 7: Since 3 = 31, we can write the equation as 3x= 31.
Step 8: Equating the exponents gives x= 1.
Step 9: Thus, the solution to the exponential equation 32x−6·3x+ 9 = 0
is x= 1.
11
Question 16
Question
Solve the exponential equation 23x−1−2x−1= 8 for x.
Solution
Step 1: Rewrite the equation using a common base. Step 2: Use the properties
of exponents to simplify the equation. Step 3: Solve for the variable y. Step 4:
Check the solution in the original equation.
Step 1: Rewrite the equation using a common base.
23x−1−2x−1= 8
Step 2: Use the properties of exponents to simplify the equation.
23x·2−1−2x·2−1= 8
23x·1
2−2x·1
2= 8
23x−1−2x−1= 8
Step 3: Solve for the variable x. Now, the equation becomes:
23x−1−2x−1= 8
Let y= 2x−1, then the equation becomes:
2y3−y= 8
2y3−y−8 = 0
Factoring the equation:
(2y+ 1)(y−2)(y+ 4) = 0
This gives us 3 possible solutions for y:y1=−1
2,y2= 2,y3=−4
Now, we substitute back to solve for x: For y1=−1
2:
2x−1=−1
2
This has no real solution since 2x−1is always positive.
For y2= 2:
2x−1= 2
x−1 = 1
x= 2
12
For y3=−4:
2x−1=−4
This has no real solution since 2x−1is always positive.
Step 4: Check the solution in the original equation. Checking x= 2:
23(2)−1−22−1= 8
25−2 = 8
32 −2 = 8
30 = 8
This solution is not valid, so there is no solution to the equation.
Question 17
Question
Solve the following exponential equation for x:32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Rewrite 27 as 33since 33= 27.
Step 2: Substitute 27 as 33into the equation 32x+1 = 27.
32x+1 = 33.
Step 3: Apply the property of exponents that states if ab=ac, then b=c.
Set 2x+ 1 = 3 to solve for x.
Step 4: Solve for x.
2x+ 1 = 3
2x= 3 −1
2x= 2
x= 1.
Step 5: Check the solution by substituting x= 1 back into the original
equation.
32(1)+1 = 33
33= 27.
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
13
Question 18
Question
A certain radioactive isotope decays at a rate of 8
Solution
Step 1: Let A(t)be the amount of the isotope remaining after thours. Since
the isotope decays at a rate of 8
A(t) = 200 ×(1 −0.08)t
Step 2: Substitute t= 12 into the formula to find the amount of the isotope
remaining after 12 hours:
A(12) = 200 ×(1 −0.08)12
Step 3: Calculate the value of A(12):
A(12) = 200 ×(0.92)12
Step 4: Use a calculator to compute the value of (0.92)12, then multiply the
result by 200 to find the amount of the isotope remaining after 12 hours.
Step 5: After performing the calculations, the amount of the isotope remain-
ing after 12 hours is approximately 102.51 grams.
Question 19
Question
A population of bacteria triples every 5 hours. Initially, there are 100 bacteria
in the population. a) Find the exponential function that models the population
size after thours. b) How many bacteria will be in the population after 20
hours?
Solution
a) Let P(t)be the population size after thours. Since the population triples
every 5 hours, we can write the exponential function as:
P(t) = 100 ·3t/5
b) To find the population after 20 hours, we substitute t= 20 into the
exponential function:
P(20) = 100 ·320/5
P(20) = 100 ·34
P(20) = 100 ·81
P(20) = 8100
After 20 hours, there will be 8100 bacteria in the population.
14
Question 20
Question
Solve the exponential equation 32x−6·3x+ 9 = 0 for x.
Solution
Step 1: Let’s rewrite 32xas (3x)2. This allows us to make a simple substitution.
Step 2: Let y= 3x. Then, the equation becomes y2−6y+ 9 = 0.
Step 3: Factor the quadratic equation y2−6y+ 9 = 0 into (y−3)2= 0.
Step 4: Solve the equation (y−3)2= 0 for y. We get y= 3.
Step 5: Substitute back y= 3xinto 3xto get 3x= 3.
Step 6: Solve the equation 3x= 3 for x. We have x= 1.
Therefore, the solution to the exponential equation 32x−6·3x+ 9 = 0 is
x= 1.
Question 21
Question
Solve the following exponential equation for x:
3x−1= 5.
Solution
Step 1: Rewrite the equation using the properties of exponents:
3x−1= 5
3x= 5 ·31
3x= 15
Step 2: Take the natural logarithm of both sides to solve for x. Remember
the property: log(a·b) = log(a) + log(b).
ln(3x) = ln(15)
xln(3) = ln(15)
x=ln(15)
ln(3)
Step 3: Use a calculator to approximate the value of x:
x≈ln(15)
ln(3) ≈2.7081
1.0986 ≈2.4655
Therefore, the solution to the equation 3x−1= 5 is x≈2.4655.
15
Question 22
Question
A population of bacteria initially has 500 bacteria and doubles every hour. How
many bacteria will there be after 4 hours?
Solution
Step 1: Let P(t)represent the population of bacteria at time tin hours. Since
the population doubles every hour, we have the exponential growth model
P(t) = 500 ·2t.
Step 2: We are interested in finding P(4), which represents the population
after 4 hours. Substitute t= 4 into the model:
P(4) = 500 ·24
Step 3: Compute 24:
24= 16
Step 4: Substitute 24= 16 back into the equation:
P(4) = 500 ·16
Step 5: Calculate the population after 4 hours:
P(4) = 500 ·16 = 8000
Step 6: Therefore, after 4 hours, there will be 8000 bacteria in the popula-
tion.
Question 23
Question
Suppose an investment grows according to the exponential model A(t) = 5000 ·
1.08t, where A(t)represents the amount of money in the account after tyears.
Find the time it takes for the investment to double.
Solution
Step 1: Set up the equation for when the investment doubles. We want to find
twhen A(t) = 2A(0).
5000 ·1.08t= 2 ·5000
1.08t= 2
Step 2: Take the natural logarithm of both sides to solve for t.
ln(1.08t)= ln(2)
16
tln(1.08) = ln(2)
Step 3: Solve for tby dividing by ln(1.08).
t=ln(2)
ln(1.08)
Step 4: Use a calculator to find the approximate value of t.
t≈ln(2)
ln(1.08) ≈0.6931
0.0792 ≈8.75
So, it takes approximately 8.75 years for the investment to double.
Question 24
Question
A certain species of bacteria doubles every 3 hours. If there are 500 bacteria to
start with, how many bacteria will there be after 10 hours? Round your answer
to the nearest whole number.
Solution
Step 1: First, we need to determine the growth rate of the bacteria. Since
the bacteria doubles every 3 hours, the growth rate can be represented by the
function N(t) = 500 ·2t/3, where N(t)is the number of bacteria after thours.
Step 2: We want to find N(10), which represents the number of bacteria
after 10 hours. Plugging in t= 10 into the function, we get:
N(10) = 500 ·210/3
Step 3: Calculating 210/3:
210/3= (21/3)10 = ( 3
√2)10 = 2.154
Step 4: Substitute 2.154 back into N(10):
N(10) = 500 ·2.154
Step 5: Calculate the final answer:
N(10) = 1077
Therefore, there will be approximately 1077 bacteria after 10 hours.
Question 25
Question
Solve the equation log3(x−1) + log3(x+ 3) = 2 for x.
17
Solution
Step 1: Apply the product rule of logarithms to combine the two logarithmic
terms on the left side:
log3((x−1)(x+ 3)) = 2
Step 2: Simplify the equation by expanding the product inside the logarithm:
log3(x2+ 2x−3) = 2
Step 3: Rewrite the logarithmic equation in exponential form:
32=x2+ 2x−3
Step 4: Simplify the exponential equation:
9 = x2+ 2x−3
Step 5: Rearrange the equation into standard quadratic form:
x2+ 2x−12 = 0
Step 6: Factor the quadratic equation:
(x+ 6)(x−2) = 0
Step 7: Set each factor equal to zero and solve for x:
x+ 6 = 0 or x−2 = 0
x=−6or x= 2
Step 8: Check both solutions for extraneous roots by substituting them back
into the original logarithmic equation. Since the logarithm of a negative number
is undefined, x=−6is an extraneous root.
Therefore, the solution to the equation log3(x−1)+log3(x+ 3) = 2 is x= 2.
Question 26
Question
Suppose the population of a city is modeled by the exponential function P(t) =
5000 ·1.02t, where P(t)is the population after tyears. Determine the initial
population and the annual growth rate of the city.
Solution
Step 1: To find the initial population, we evaluate P(0).
P(0) = 5000 ·1.020= 5000 ·1 = 5000
Therefore, the initial population of the city is 5000.
Step 2: The annual growth rate is the constant multiplier in the exponential
function. In this case, the growth rate is 1.02, which means the city’s population
grows by 2% annually.
18
Question 27
Question
A certain species of bacteria doubles in population every 6 hours. If there are
initially 100 bacteria in the population, how many bacteria will there be after
24 hours?
Solution
Step 1: To find the growth factor of the bacteria population, we use the formula
A=P(1 + r)t, where: - Ais the final amount, - Pis the initial amount, - ris
the growth rate, and - tis the time in hours.
Step 2: Since the bacteria doubles every 6 hours, the growth rate is r= 1,
meaning the bacteria population grows by 100
Step 3: Substituting P= 100,r= 1, and t= 24 into the formula, we have:
A= 100(1 + 1)24
Step 4: Simplify the expression inside the parentheses:
A= 100(2)24
Step 5: Calculate 224:
A= 100 ×16,777,216
Step 6: Multiply to find the final amount of bacteria:
A= 1,677,721,600
Therefore, after 24 hours, there will be 1,677,721,600 bacteria in the popu-
lation.
Question 28
Question
Solve the exponential equation for x:3x+2 −3x−1= 20.
Solution
To solve the given exponential equation, we can use the properties of exponents
to manipulate the equation so that we can solve for x.
Step 1: Rewrite the equation using properties of exponents. We
can rewrite the equation as 3x·32−3x·3−1= 20.
Step 2: Simplify the equation. Simplifying the equation gives 9·3x−3x
3=
20.
19
Step 3: Combine like terms. Combining like terms yields 8·3x= 20.
Step 4: Solve for x.To solve for x, we first divide both sides by 8:
3x=20
8=5
2.
Step 5: Take the logarithm of both sides. Taking the natural logarithm
of both sides gives ln(3x) = ln (5
2).
Step 6: Apply the power rule of logarithms. Using the power rule, we
have xln(3) = ln (5
2).
Step 7: Solve for x.Finally, we can solve for xby dividing by ln(3):
x=ln(5
2)
ln(3) .
Therefore, the solution to the exponential equation 3x+2 −3x−1= 20 is
x=ln(5
2)
ln(3) .
Question 29
Question
Let f(x) = 2x+ 3 and g(x) = log2(x−1). Find the value(s) of xfor which
f(x) = g(x).
Solution
Step 1: Set f(x)equal to g(x)and solve for x.
2x+ 3 = log2(x−1)
2x+ 3 = ln(x−1)
ln(2) (Using change of base formula)
2x+ 3 = ln(x−1)
ln(2)
2x=ln(x−1)
ln(2) −3
20
Step 2: Convert the logarithmic expression to exponential form.
2x=ln(x−1)
ln(2) −3
2x= 2 (1
x−1)−3
2x=2
x−1−3
2x=2−3(x−1)
x−1
2x=2−3x+ 3
x−1
2x=5−3x
x−1
Step 3: Write the equation as a single logarithmic function.
2x=5−3x
x−1
2x=5−3x
x−1
log2(2x) = log2(5−3x
x−1)
x= log2(5−3x
x−1)
Therefore, the value of xfor which f(x) = g(x)is x= log2(5−3x
x−1).
Question 30
Question
A bacteria culture starts with 1000 bacteria and doubles in size every hour.
Write an exponential function to model the population of the bacteria after t
hours. Then, find the population after 5 hours and round to the nearest whole
number.
Solution
Step 1: Let P(t)be the population of the bacteria after thours. Since the
bacteria doubles in size every hour, we can write the exponential function as
P(t) = 1000 ·2t.
21
Step 2: To find the population after 5 hours, we substitute t= 5 into the
exponential function:
P(5) = 1000 ·25
P(5) = 1000 ·32
P(5) = 32000
Therefore, the population of the bacteria after 5 hours is 32,000.
Question 31
Question
Solve the exponential equation: 2x−3−2x−2= 12.
Solution
Step 1: Rewrite the equation using a common base. Step 2: Apply properties of
exponents to simplify the equation. Step 3: Solve the resulting linear equation
for 2x. Step 4: Determine the value of xusing the logarithmic function. Step
5: Check the solution for extraneous roots.
Step 1: Rewrite the equation using a common base. We can rewrite the
equation as 2x−3−2x−3·2 = 12.
Step 2: Apply properties of exponents to simplify the equation. Simplifying
further, we have 2x−3(1 −2) = 12, which simplifies to −2·2x−3= 12.
Step 3: Solve the resulting linear equation for 2x. Dividing both sides by
−2gives 2x−3=−6.
Step 4: Determine the value of xusing the logarithmic function. Taking
the logarithm of both sides, we get log(2x−3)= log(−6). Using the property
log(ab)=blog(a)and the fact that the logarithm of a negative number is
undefined, we conclude that the equation has no solution.
Step 5: Check the solution for extraneous roots. Since the logarithm of a
negative number is undefined, there are no real solutions to the original equation.
Question 32
Question
Solve the exponential equation 3x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3.
Step 2: 27 = 33, since 27 = 3 ·3·3.
Step 3: Substitute 33back into the original equation and solve for x.
Step 4: 3x= 33
22
Step 5: Since the bases are the same, set the exponents equal to each other.
Step 6: x= 3
Therefore the solution to the exponential equation 3x= 27 is x= 3.
Question 33
Question
A population of bacteria starts with 1000 bacteria and triples every 4 hours.
Write an exponential equation representing the population of the bacteria as a
function of time, and determine how many bacteria there will be after 12 hours.
Solution
Step 1: Let P(t)represent the population of bacteria at time t. Since the
population triples every 4 hours, we have exponential growth with a growth
factor of 3. The initial population is 1000 bacteria. Thus, the exponential
equation is given by:
P(t) = 1000 ×3t/4
Step 2: To find the population after 12 hours, substitute t= 12 into the
equation and simplify:
P(12) = 1000 ×312/4= 1000 ×33= 1000 ×27 = 27000
Therefore, after 12 hours, there will be 27,000 bacteria in the population.
Question 34
Question
Solve for x:2x+1 + 2x= 24.
Solution
Step 1: Simplify the left side of the equation by factoring out a common factor.
2x+1 + 2x= 2 ·2x+ 2x
= 2 ·2x+ 1 ·2x
= (2 + 1) ·2x
= 3 ·2x
23
Step 2: Substitute the simplified expression back into the equation and solve
for x.
3·2x= 24
2x=24
3
2x= 8
Step 3: Rewrite 8 as a power of 2 to help solve for x.
2x= 23
Step 4: Since the bases are the same, the exponents must be equal.
x= 3
Therefore, the solution to the equation 2x+1 + 2x= 24 is x= 3.
Question 35
Question
Solve the exponential equation 3x+1
27 = 9.
Solution
Step 1: Rewrite both sides of the equation with the same base:
3x+1
27 = 32
Step 2: Rewrite 27 as 33:
3x+1
33= 32
Step 3: Apply the quotient rule of exponents to simplify the left side:
3x+1−3= 32
Step 4: Simplify the exponent on the left side:
3x−2= 32
Step 5: Since the bases are the same, set the exponents equal to each other:
x−2 = 2
Step 6: Solve for x:
x= 2 + 2
Step 7: Simplify to find the final solution:
x= 4
Therefore, the solution to the exponential equation 3x+1
27 = 9 is x= 4.
24
Step 6: Divide by 5to solve for 3x:
3x=9
5
Step 7: Rewrite 9
5as 3log3
9
5:
3x= 3log3
9
5
Step 8: Since the bases are the same, we can set the exponents equal to each
other:
x= log3
9
5
Therefore, the solution to the exponential equation 3x−1−2·3x+ 3 = 0 is
x= log39
5.
Question 2
Question
Solve the equation 3x−1= 27 for x.
Solution
Step 1: Write 27 as a power of 3:
27 = 33
Step 2: Substitute 27 as 33in the equation and solve for x:
3x−1= 33
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 adding 1to both sides:
x= 4
Step 5: Thus, the solution to the equation 3x−1= 27 is x= 4.
Question 3
Question
Samantha invests $5000 in a savings account that pays an annual interest rate
of 4.5%, compounded continuously. How long will it take for her investment to
double?
2
Solution
Step 1: We can model Samantha’s investment using the continuous compound
interest formula:
A=P·ert
where: - Ais the amount after time t, - Pis the principal amount (initial
investment), - ris the annual interest rate, - tis the time in years, and - eis
the base of the natural logarithm.
Step 2: Since Samantha wants to double her investment, the amount after
doubling will be $10,000. Thus, we have:
10000 = 5000 ·e0.045t
Step 3: Divide both sides by 5000 to solve for the exponential equation:
2 = e0.045t
Step 4: To solve for t, we take the natural logarithm of both sides of the
equation:
ln(2) = ln(e0.045t)
Step 5: Use the property of logarithms to bring down the exponent:
ln(2) = 0.045tln(e)
Step 6: Recall that ln(e) = 1, so our equation simplifies to:
ln(2) = 0.045t
Step 7: Now, solve for tby dividing both sides by 0.045:
t=ln(2)
0.045
Step 8: Use a calculator to approximate the value of t:
t≈ln(2)
0.045 ≈0.6931
0.045 ≈15.47
Step 9: Therefore, it will take approximately 15.47 years for Samantha’s
investment to double.
Question 4
Question
Solve the exponential equation: 3x−27 = 0.
3
Solution
Step 1: Add 27 to both sides of the equation to isolate the exponential term:
3x= 27
Step 2: Rewrite 27 as 33:
3x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
x= 3
Step 4: Therefore, the solution to the exponential equation 3x−27 = 0 is
x= 3.
Question 5
Question
Samantha invested $10,000 in a savings account that pays an annual interest
rate of 4.5%, compounded continuously. How long will it take for her investment
to double in value?
Solution
Let Pbe the initial investment of $10,000, rbe the annual interest rate (4.5%
or 0.045), and tbe the time in years it takes for the investment to double.
We can use the continuous compound interest formula to model the situation:
A=P ert
Since Samantha’s investment needs to double, the final amount Awill be 2P:
2P=P e0.045t
Dividing both sides by P, we get:
2 = e0.045t
To solve for t, we take the natural logarithm of both sides:
ln 2 = ln e0.045t
ln 2 = 0.045t
t=ln 2
0.045
Step 1: Calculate ln 2.
ln 2 ≈0.693
4
Step 2: Plug ln 2 into the formula to find t.
t=0.693
0.045
t≈15.4years
Therefore, it will take approximately 15.4 years for Samantha’s investment
to double in value when compounded continuously.
Question 6
Question
Samantha invested $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much will Samantha have in the account after 10
years?
Solution
Step 1: To find the amount of money in the account after 10 years, we can use
the continuous compounding formula:
A=P·ert
where: - Ais the amount of money in the account after tyears, - Pis the
principal amount initially invested ($5000 in this case), - ris the annual interest
rate (3.5% or 0.035 as a decimal), - tis the time the money is invested for (10
years), and - eis the base of the natural logarithm (approximately 2.71828).
Step 2: Substituting the given values into the formula, we get:
A= 5000 ·e0.035·10
Step 3: Calculate the exponential term:
A≈5000 ·e0.35
Step 4: Evaluate the exponential term:
A≈5000 ·1.4203
Step 5: Multiply to find the final amount:
A≈$7101.50
Therefore, Samantha will have approximately $7101.50 in the account after
10 years.
5
Question 7
Question
Solve for x:5x−1= 125
Solution
Step 1: Rewrite 125 as a power of 5.
125 = 53
Step 2: Substitute 125 with 53in the equation.
5x−1= 53
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
x−1 = 3
Step 4: Add 1 to both sides to solve for x.
x= 4
Thus, the solution to the equation is x= 4.
Question 8
Question
Solve the exponential equation: 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 = 33into the equation 32x+1 = 27.
32x+1 = 33
Step 3: Apply the property of equality for exponents, am=animplies
m=n, to solve for x+ 1.
2x+ 1 = 3
6
Step 4: Solve for xby isolating x.
2x= 3 −1
2x= 2
x= 1
Step 5: Finally, verify the solution x= 1 by substituting it back into the
original equation.
32(1)+1 = 27
33= 27
27 = 27
Thus, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 9
Question
Solve the exponential equation 2x+1 = 8.
Solution
Step 1: Rewrite 8 as a power of 2.
Since 8 = 23,we have 2x+1 = 23.
Step 2: Set the exponents equal to each other.
Therefore, x+ 1 = 3.
Step 3: Solve for x.
Subtracting 1 from both sides gives us x= 2.
Step 4: Verify the solution.
Substitute x= 2 back into the original equation: 22+1 = 23= 8,
which confirms that x= 2 is the correct solution.
Question 10
Question
Samantha invested $5000 in an account earning 3.5% annual interest com-
pounded continuously. How much will she have in the account after 10 years?
7
Solution
Step 1: The formula for compound interest compounded continuously is given
by A=P·ert, where: - Ais the amount of money accumulated after tyears,
-Pis the principal amount (the initial amount of money), - ris the annual
interest rate (written as a decimal), - tis the time the money is invested for (in
years), and - eis the base of the natural logarithm.
Step 2: Given that the principal amount P= $5000, the interest rate r=
0.035 (3.5% written as a decimal), and the time t= 10 years, we can plug these
values into the formula:
A= 5000 ·e0.035·10
Step 3: Simplifying the formula:
A= 5000 ·e0.35
Step 4: Calculating e0.35:
A≈5000 ·1.419067
Step 5: Finding the final amount in the account after 10 years:
A≈7095.335
Therefore, after 10 years, Samantha will have approximately $7095.34 in the
account.
Question 11
Question
Suzanne invests $200 in a savings account that pays 4% interest compounded
quarterly. How much money will be in the account after 5 years?
Solution
Step 1: First, we need to determine the annual interest rate (r) and the number
of compounding periods per year (n). The formula for compound interest is:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= annual interest rate (as a decimal), n= number
of compounding periods per year, and t= time in years. Given that the interest
is compounded quarterly, n= 4 and r= 0.04.
Step 2: Let’s substitute the values into the formula and calculate the amount
of money in the account after 5 years:
A= 200 (1 + 0.04
4)4·5
8
A= 200 (1 + 0.01)20
A= 200 ×1.0120
Step 3: Now, we compute 1.0120 to find the final amount in the account
after 5 years.
A≈200 ×1.221386
A≈244.2772
Therefore, after 5 years, Suzanne will have approximately $244.28 in the
savings account.
Question 12
Question
A certain amount of money is invested at an annual interest rate of 5
Solution
Step 1: Given that the initial investment doubles in 10 years, we can express
this information using the continuous compound interest formula:
A=P·ert
where: - Ais the final amount after time t, - Pis the initial amount invested,
-ris the annual interest rate (decimal form), and - tis the time the money is
invested for.
Step 2: Since the initial investment doubles, the final amount after 10 years
is twice the initial amount:
2P=P·e0.05·10
Step 3: Simplify the equation from Step 2 to solve for P:
2 = e0.5
Step 4: Take natural logarithm (ln) of both sides of the equation to eliminate
the exponential:
ln(2) = ln(e0.5)
Step 5: Use the property ln(ex) = xto simplify the equation from Step 4:
ln(2) = 0.5
Step 6: Solve for Pby converting the equation in Step 5 back to exponential
form:
e0.5= 2
Step 7: Therefore, the initial amount of money invested was 2.
9
Question 13
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4% compounded continuously. How much will her investment be worth after
10 years?
Solution
Step 1: Identify the variables and known values in the problem. Let’s denote
the initial amount invested as P= $5000, the annual interest rate as r= 0.04,
the number of years as t= 10, and the amount after tyears as A.
Step 2: Use the formula for continuous compounding to find the future value
of the investment. The formula for continuous compounding is given by:
A=P·ert
Step 3: Substitute the values into the formula and solve for A. Substitute
P= 5000,r= 0.04, and t= 10 into the formula:
A= 5000 ·e0.04·10
Step 4: Calculate the value of A.
A= 5000 ·e0.4
A= 5000 ·e0.4≈5000 ·1.4918 ≈7459.08
Therefore, after 10 years, Samantha’s investment will be worth approxi-
mately $7459.08.
Question 14
Question
Samantha invested some money in a savings account that earns an annual inter-
est rate of 6.5%, compounded quarterly. If she initially invested $10,000, how
much money will she have in the account after 5 years?
Solution
Step 1: Identify the relevant compound interest formula. The formula for com-
pound interest is given by:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after n years, including interest,
P= the principal amount (the initial amount of money), r= annual interest
10
rate (decimal), n= number of times that interest is compounded per year, t=
time the money is invested for in years.
Step 2: Plug in the known values into the formula. In this case, we have:
P= $10,000 (initial investment), r= 6.5% = 0.065 (annual interest rate in
decimal form), n= 4 (interest is compounded quarterly), t= 5 years.
Step 3: Calculate the amount of money Samantha will have in the account
after 5 years using the compound interest formula:
A= 10000 (1 + 0.065
4)4×5
Step 4: Simplify the expression inside the parentheses first:
1 + 0.065
4= 1 + 0.01625 = 1.01625
Step 5: Plug this back into the formula and calculate:
A= 10000 ×(1.01625)20
Step 6: Evaluate (1.01625)20:
(1.01625)20 ≈1.3548
Step 7: Finally, calculate the total amount of money Samantha will have
after 5 years:
A= 10000 ×1.3548 ≈$13,548
Therefore, after 5 years, Samantha will have approximately $13,548 in her
savings account.
Question 15
Question
Solve the exponential equation 32x−6·3x+ 9 = 0.
Solution
Step 1: Let’s make a substitution to simplify the equation. Let y= 3x. Then,
the original equation becomes y2−6y+ 9 = 0.
Step 2: Now, we need to solve the quadratic equation y2−6y+ 9 = 0.
Step 3: Factoring the quadratic equation gives us (y−3)2= 0.
Step 4: To find the value of y, we take the square root of both sides: y−3 = 0.
Step 5: Therefore, y= 3.
Step 6: Now, substitute back 3xfor y:3x= 3.
Step 7: Since 3 = 31, we can write the equation as 3x= 31.
Step 8: Equating the exponents gives x= 1.
Step 9: Thus, the solution to the exponential equation 32x−6·3x+ 9 = 0
is x= 1.
11
Question 16
Question
Solve the exponential equation 23x−1−2x−1= 8 for x.
Solution
Step 1: Rewrite the equation using a common base. Step 2: Use the properties
of exponents to simplify the equation. Step 3: Solve for the variable y. Step 4:
Check the solution in the original equation.
Step 1: Rewrite the equation using a common base.
23x−1−2x−1= 8
Step 2: Use the properties of exponents to simplify the equation.
23x·2−1−2x·2−1= 8
23x·1
2−2x·1
2= 8
23x−1−2x−1= 8
Step 3: Solve for the variable x. Now, the equation becomes:
23x−1−2x−1= 8
Let y= 2x−1, then the equation becomes:
2y3−y= 8
2y3−y−8 = 0
Factoring the equation:
(2y+ 1)(y−2)(y+ 4) = 0
This gives us 3 possible solutions for y:y1=−1
2,y2= 2,y3=−4
Now, we substitute back to solve for x: For y1=−1
2:
2x−1=−1
2
This has no real solution since 2x−1is always positive.
For y2= 2:
2x−1= 2
x−1 = 1
x= 2
12
For y3=−4:
2x−1=−4
This has no real solution since 2x−1is always positive.
Step 4: Check the solution in the original equation. Checking x= 2:
23(2)−1−22−1= 8
25−2 = 8
32 −2 = 8
30 = 8
This solution is not valid, so there is no solution to the equation.
Question 17
Question
Solve the following exponential equation for x:32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Rewrite 27 as 33since 33= 27.
Step 2: Substitute 27 as 33into the equation 32x+1 = 27.
32x+1 = 33.
Step 3: Apply the property of exponents that states if ab=ac, then b=c.
Set 2x+ 1 = 3 to solve for x.
Step 4: Solve for x.
2x+ 1 = 3
2x= 3 −1
2x= 2
x= 1.
Step 5: Check the solution by substituting x= 1 back into the original
equation.
32(1)+1 = 33
33= 27.
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
13
Question 18
Question
A certain radioactive isotope decays at a rate of 8
Solution
Step 1: Let A(t)be the amount of the isotope remaining after thours. Since
the isotope decays at a rate of 8
A(t) = 200 ×(1 −0.08)t
Step 2: Substitute t= 12 into the formula to find the amount of the isotope
remaining after 12 hours:
A(12) = 200 ×(1 −0.08)12
Step 3: Calculate the value of A(12):
A(12) = 200 ×(0.92)12
Step 4: Use a calculator to compute the value of (0.92)12, then multiply the
result by 200 to find the amount of the isotope remaining after 12 hours.
Step 5: After performing the calculations, the amount of the isotope remain-
ing after 12 hours is approximately 102.51 grams.
Question 19
Question
A population of bacteria triples every 5 hours. Initially, there are 100 bacteria
in the population. a) Find the exponential function that models the population
size after thours. b) How many bacteria will be in the population after 20
hours?
Solution
a) Let P(t)be the population size after thours. Since the population triples
every 5 hours, we can write the exponential function as:
P(t) = 100 ·3t/5
b) To find the population after 20 hours, we substitute t= 20 into the
exponential function:
P(20) = 100 ·320/5
P(20) = 100 ·34
P(20) = 100 ·81
P(20) = 8100
After 20 hours, there will be 8100 bacteria in the population.
14
Question 20
Question
Solve the exponential equation 32x−6·3x+ 9 = 0 for x.
Solution
Step 1: Let’s rewrite 32xas (3x)2. This allows us to make a simple substitution.
Step 2: Let y= 3x. Then, the equation becomes y2−6y+ 9 = 0.
Step 3: Factor the quadratic equation y2−6y+ 9 = 0 into (y−3)2= 0.
Step 4: Solve the equation (y−3)2= 0 for y. We get y= 3.
Step 5: Substitute back y= 3xinto 3xto get 3x= 3.
Step 6: Solve the equation 3x= 3 for x. We have x= 1.
Therefore, the solution to the exponential equation 32x−6·3x+ 9 = 0 is
x= 1.
Question 21
Question
Solve the following exponential equation for x:
3x−1= 5.
Solution
Step 1: Rewrite the equation using the properties of exponents:
3x−1= 5
3x= 5 ·31
3x= 15
Step 2: Take the natural logarithm of both sides to solve for x. Remember
the property: log(a·b) = log(a) + log(b).
ln(3x) = ln(15)
xln(3) = ln(15)
x=ln(15)
ln(3)
Step 3: Use a calculator to approximate the value of x:
x≈ln(15)
ln(3) ≈2.7081
1.0986 ≈2.4655
Therefore, the solution to the equation 3x−1= 5 is x≈2.4655.
15
Question 22
Question
A population of bacteria initially has 500 bacteria and doubles every hour. How
many bacteria will there be after 4 hours?
Solution
Step 1: Let P(t)represent the population of bacteria at time tin hours. Since
the population doubles every hour, we have the exponential growth model
P(t) = 500 ·2t.
Step 2: We are interested in finding P(4), which represents the population
after 4 hours. Substitute t= 4 into the model:
P(4) = 500 ·24
Step 3: Compute 24:
24= 16
Step 4: Substitute 24= 16 back into the equation:
P(4) = 500 ·16
Step 5: Calculate the population after 4 hours:
P(4) = 500 ·16 = 8000
Step 6: Therefore, after 4 hours, there will be 8000 bacteria in the popula-
tion.
Question 23
Question
Suppose an investment grows according to the exponential model A(t) = 5000 ·
1.08t, where A(t)represents the amount of money in the account after tyears.
Find the time it takes for the investment to double.
Solution
Step 1: Set up the equation for when the investment doubles. We want to find
twhen A(t) = 2A(0).
5000 ·1.08t= 2 ·5000
1.08t= 2
Step 2: Take the natural logarithm of both sides to solve for t.
ln(1.08t)= ln(2)
16
tln(1.08) = ln(2)
Step 3: Solve for tby dividing by ln(1.08).
t=ln(2)
ln(1.08)
Step 4: Use a calculator to find the approximate value of t.
t≈ln(2)
ln(1.08) ≈0.6931
0.0792 ≈8.75
So, it takes approximately 8.75 years for the investment to double.
Question 24
Question
A certain species of bacteria doubles every 3 hours. If there are 500 bacteria to
start with, how many bacteria will there be after 10 hours? Round your answer
to the nearest whole number.
Solution
Step 1: First, we need to determine the growth rate of the bacteria. Since
the bacteria doubles every 3 hours, the growth rate can be represented by the
function N(t) = 500 ·2t/3, where N(t)is the number of bacteria after thours.
Step 2: We want to find N(10), which represents the number of bacteria
after 10 hours. Plugging in t= 10 into the function, we get:
N(10) = 500 ·210/3
Step 3: Calculating 210/3:
210/3= (21/3)10 = ( 3
√2)10 = 2.154
Step 4: Substitute 2.154 back into N(10):
N(10) = 500 ·2.154
Step 5: Calculate the final answer:
N(10) = 1077
Therefore, there will be approximately 1077 bacteria after 10 hours.
Question 25
Question
Solve the equation log3(x−1) + log3(x+ 3) = 2 for x.
17
Solution
Step 1: Apply the product rule of logarithms to combine the two logarithmic
terms on the left side:
log3((x−1)(x+ 3)) = 2
Step 2: Simplify the equation by expanding the product inside the logarithm:
log3(x2+ 2x−3) = 2
Step 3: Rewrite the logarithmic equation in exponential form:
32=x2+ 2x−3
Step 4: Simplify the exponential equation:
9 = x2+ 2x−3
Step 5: Rearrange the equation into standard quadratic form:
x2+ 2x−12 = 0
Step 6: Factor the quadratic equation:
(x+ 6)(x−2) = 0
Step 7: Set each factor equal to zero and solve for x:
x+ 6 = 0 or x−2 = 0
x=−6or x= 2
Step 8: Check both solutions for extraneous roots by substituting them back
into the original logarithmic equation. Since the logarithm of a negative number
is undefined, x=−6is an extraneous root.
Therefore, the solution to the equation log3(x−1)+log3(x+ 3) = 2 is x= 2.
Question 26
Question
Suppose the population of a city is modeled by the exponential function P(t) =
5000 ·1.02t, where P(t)is the population after tyears. Determine the initial
population and the annual growth rate of the city.
Solution
Step 1: To find the initial population, we evaluate P(0).
P(0) = 5000 ·1.020= 5000 ·1 = 5000
Therefore, the initial population of the city is 5000.
Step 2: The annual growth rate is the constant multiplier in the exponential
function. In this case, the growth rate is 1.02, which means the city’s population
grows by 2% annually.
18
Question 27
Question
A certain species of bacteria doubles in population every 6 hours. If there are
initially 100 bacteria in the population, how many bacteria will there be after
24 hours?
Solution
Step 1: To find the growth factor of the bacteria population, we use the formula
A=P(1 + r)t, where: - Ais the final amount, - Pis the initial amount, - ris
the growth rate, and - tis the time in hours.
Step 2: Since the bacteria doubles every 6 hours, the growth rate is r= 1,
meaning the bacteria population grows by 100
Step 3: Substituting P= 100,r= 1, and t= 24 into the formula, we have:
A= 100(1 + 1)24
Step 4: Simplify the expression inside the parentheses:
A= 100(2)24
Step 5: Calculate 224:
A= 100 ×16,777,216
Step 6: Multiply to find the final amount of bacteria:
A= 1,677,721,600
Therefore, after 24 hours, there will be 1,677,721,600 bacteria in the popu-
lation.
Question 28
Question
Solve the exponential equation for x:3x+2 −3x−1= 20.
Solution
To solve the given exponential equation, we can use the properties of exponents
to manipulate the equation so that we can solve for x.
Step 1: Rewrite the equation using properties of exponents. We
can rewrite the equation as 3x·32−3x·3−1= 20.
Step 2: Simplify the equation. Simplifying the equation gives 9·3x−3x
3=
20.
19
Step 3: Combine like terms. Combining like terms yields 8·3x= 20.
Step 4: Solve for x.To solve for x, we first divide both sides by 8:
3x=20
8=5
2.
Step 5: Take the logarithm of both sides. Taking the natural logarithm
of both sides gives ln(3x) = ln (5
2).
Step 6: Apply the power rule of logarithms. Using the power rule, we
have xln(3) = ln (5
2).
Step 7: Solve for x.Finally, we can solve for xby dividing by ln(3):
x=ln(5
2)
ln(3) .
Therefore, the solution to the exponential equation 3x+2 −3x−1= 20 is
x=ln(5
2)
ln(3) .
Question 29
Question
Let f(x) = 2x+ 3 and g(x) = log2(x−1). Find the value(s) of xfor which
f(x) = g(x).
Solution
Step 1: Set f(x)equal to g(x)and solve for x.
2x+ 3 = log2(x−1)
2x+ 3 = ln(x−1)
ln(2) (Using change of base formula)
2x+ 3 = ln(x−1)
ln(2)
2x=ln(x−1)
ln(2) −3
20
Step 2: Convert the logarithmic expression to exponential form.
2x=ln(x−1)
ln(2) −3
2x= 2 (1
x−1)−3
2x=2
x−1−3
2x=2−3(x−1)
x−1
2x=2−3x+ 3
x−1
2x=5−3x
x−1
Step 3: Write the equation as a single logarithmic function.
2x=5−3x
x−1
2x=5−3x
x−1
log2(2x) = log2(5−3x
x−1)
x= log2(5−3x
x−1)
Therefore, the value of xfor which f(x) = g(x)is x= log2(5−3x
x−1).
Question 30
Question
A bacteria culture starts with 1000 bacteria and doubles in size every hour.
Write an exponential function to model the population of the bacteria after t
hours. Then, find the population after 5 hours and round to the nearest whole
number.
Solution
Step 1: Let P(t)be the population of the bacteria after thours. Since the
bacteria doubles in size every hour, we can write the exponential function as
P(t) = 1000 ·2t.
21
Step 2: To find the population after 5 hours, we substitute t= 5 into the
exponential function:
P(5) = 1000 ·25
P(5) = 1000 ·32
P(5) = 32000
Therefore, the population of the bacteria after 5 hours is 32,000.
Question 31
Question
Solve the exponential equation: 2x−3−2x−2= 12.
Solution
Step 1: Rewrite the equation using a common base. Step 2: Apply properties of
exponents to simplify the equation. Step 3: Solve the resulting linear equation
for 2x. Step 4: Determine the value of xusing the logarithmic function. Step
5: Check the solution for extraneous roots.
Step 1: Rewrite the equation using a common base. We can rewrite the
equation as 2x−3−2x−3·2 = 12.
Step 2: Apply properties of exponents to simplify the equation. Simplifying
further, we have 2x−3(1 −2) = 12, which simplifies to −2·2x−3= 12.
Step 3: Solve the resulting linear equation for 2x. Dividing both sides by
−2gives 2x−3=−6.
Step 4: Determine the value of xusing the logarithmic function. Taking
the logarithm of both sides, we get log(2x−3)= log(−6). Using the property
log(ab)=blog(a)and the fact that the logarithm of a negative number is
undefined, we conclude that the equation has no solution.
Step 5: Check the solution for extraneous roots. Since the logarithm of a
negative number is undefined, there are no real solutions to the original equation.
Question 32
Question
Solve the exponential equation 3x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3.
Step 2: 27 = 33, since 27 = 3 ·3·3.
Step 3: Substitute 33back into the original equation and solve for x.
Step 4: 3x= 33
22
Step 5: Since the bases are the same, set the exponents equal to each other.
Step 6: x= 3
Therefore the solution to the exponential equation 3x= 27 is x= 3.
Question 33
Question
A population of bacteria starts with 1000 bacteria and triples every 4 hours.
Write an exponential equation representing the population of the bacteria as a
function of time, and determine how many bacteria there will be after 12 hours.
Solution
Step 1: Let P(t)represent the population of bacteria at time t. Since the
population triples every 4 hours, we have exponential growth with a growth
factor of 3. The initial population is 1000 bacteria. Thus, the exponential
equation is given by:
P(t) = 1000 ×3t/4
Step 2: To find the population after 12 hours, substitute t= 12 into the
equation and simplify:
P(12) = 1000 ×312/4= 1000 ×33= 1000 ×27 = 27000
Therefore, after 12 hours, there will be 27,000 bacteria in the population.
Question 34
Question
Solve for x:2x+1 + 2x= 24.
Solution
Step 1: Simplify the left side of the equation by factoring out a common factor.
2x+1 + 2x= 2 ·2x+ 2x
= 2 ·2x+ 1 ·2x
= (2 + 1) ·2x
= 3 ·2x
23
Step 2: Substitute the simplified expression back into the equation and solve
for x.
3·2x= 24
2x=24
3
2x= 8
Step 3: Rewrite 8 as a power of 2 to help solve for x.
2x= 23
Step 4: Since the bases are the same, the exponents must be equal.
x= 3
Therefore, the solution to the equation 2x+1 + 2x= 24 is x= 3.
Question 35
Question
Solve the exponential equation 3x+1
27 = 9.
Solution
Step 1: Rewrite both sides of the equation with the same base:
3x+1
27 = 32
Step 2: Rewrite 27 as 33:
3x+1
33= 32
Step 3: Apply the quotient rule of exponents to simplify the left side:
3x+1−3= 32
Step 4: Simplify the exponent on the left side:
3x−2= 32
Step 5: Since the bases are the same, set the exponents equal to each other:
x−2 = 2
Step 6: Solve for x:
x= 2 + 2
Step 7: Simplify to find the final solution:
x= 4
Therefore, the solution to the exponential equation 3x+1
27 = 9 is x= 4.
24
Step 6: Divide by 5to solve for 3x:
3x=9
5
Step 7: Rewrite 9
5as 3log3
9
5:
3x= 3log3
9
5
Step 8: Since the bases are the same, we can set the exponents equal to each
other:
x= log3
9
5
Therefore, the solution to the exponential equation 3x−1−2·3x+ 3 = 0 is
x= log39
5.
Question 2
Question
Solve the equation 3x−1= 27 for x.
Solution
Step 1: Write 27 as a power of 3:
27 = 33
Step 2: Substitute 27 as 33in the equation and solve for x:
3x−1= 33
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 adding 1to both sides:
x= 4
Step 5: Thus, the solution to the equation 3x−1= 27 is x= 4.
Question 3
Question
Samantha invests $5000 in a savings account that pays an annual interest rate
of 4.5%, compounded continuously. How long will it take for her investment to
double?
2
Solution
Step 1: We can model Samantha’s investment using the continuous compound
interest formula:
A=P·ert
where: - Ais the amount after time t, - Pis the principal amount (initial
investment), - ris the annual interest rate, - tis the time in years, and - eis
the base of the natural logarithm.
Step 2: Since Samantha wants to double her investment, the amount after
doubling will be $10,000. Thus, we have:
10000 = 5000 ·e0.045t
Step 3: Divide both sides by 5000 to solve for the exponential equation:
2 = e0.045t
Step 4: To solve for t, we take the natural logarithm of both sides of the
equation:
ln(2) = ln(e0.045t)
Step 5: Use the property of logarithms to bring down the exponent:
ln(2) = 0.045tln(e)
Step 6: Recall that ln(e) = 1, so our equation simplifies to:
ln(2) = 0.045t
Step 7: Now, solve for tby dividing both sides by 0.045:
t=ln(2)
0.045
Step 8: Use a calculator to approximate the value of t:
t≈ln(2)
0.045 ≈0.6931
0.045 ≈15.47
Step 9: Therefore, it will take approximately 15.47 years for Samantha’s
investment to double.
Question 4
Question
Solve the exponential equation: 3x−27 = 0.
3
Solution
Step 1: Add 27 to both sides of the equation to isolate the exponential term:
3x= 27
Step 2: Rewrite 27 as 33:
3x= 33
Step 3: Since the bases are the same, set the exponents equal to each other:
x= 3
Step 4: Therefore, the solution to the exponential equation 3x−27 = 0 is
x= 3.
Question 5
Question
Samantha invested $10,000 in a savings account that pays an annual interest
rate of 4.5%, compounded continuously. How long will it take for her investment
to double in value?
Solution
Let Pbe the initial investment of $10,000, rbe the annual interest rate (4.5%
or 0.045), and tbe the time in years it takes for the investment to double.
We can use the continuous compound interest formula to model the situation:
A=P ert
Since Samantha’s investment needs to double, the final amount Awill be 2P:
2P=P e0.045t
Dividing both sides by P, we get:
2 = e0.045t
To solve for t, we take the natural logarithm of both sides:
ln 2 = ln e0.045t
ln 2 = 0.045t
t=ln 2
0.045
Step 1: Calculate ln 2.
ln 2 ≈0.693
4
Step 2: Plug ln 2 into the formula to find t.
t=0.693
0.045
t≈15.4years
Therefore, it will take approximately 15.4 years for Samantha’s investment
to double in value when compounded continuously.
Question 6
Question
Samantha invested $5000 in a savings account that earns 3.5% interest com-
pounded continuously. How much will Samantha have in the account after 10
years?
Solution
Step 1: To find the amount of money in the account after 10 years, we can use
the continuous compounding formula:
A=P·ert
where: - Ais the amount of money in the account after tyears, - Pis the
principal amount initially invested ($5000 in this case), - ris the annual interest
rate (3.5% or 0.035 as a decimal), - tis the time the money is invested for (10
years), and - eis the base of the natural logarithm (approximately 2.71828).
Step 2: Substituting the given values into the formula, we get:
A= 5000 ·e0.035·10
Step 3: Calculate the exponential term:
A≈5000 ·e0.35
Step 4: Evaluate the exponential term:
A≈5000 ·1.4203
Step 5: Multiply to find the final amount:
A≈$7101.50
Therefore, Samantha will have approximately $7101.50 in the account after
10 years.
5
Question 7
Question
Solve for x:5x−1= 125
Solution
Step 1: Rewrite 125 as a power of 5.
125 = 53
Step 2: Substitute 125 with 53in the equation.
5x−1= 53
Step 3: Since the bases are the same, we can set the exponents equal to each
other.
x−1 = 3
Step 4: Add 1 to both sides to solve for x.
x= 4
Thus, the solution to the equation is x= 4.
Question 8
Question
Solve the exponential equation: 32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
27 = 33
Step 2: Substitute 27 = 33into the equation 32x+1 = 27.
32x+1 = 33
Step 3: Apply the property of equality for exponents, am=animplies
m=n, to solve for x+ 1.
2x+ 1 = 3
6
Step 4: Solve for xby isolating x.
2x= 3 −1
2x= 2
x= 1
Step 5: Finally, verify the solution x= 1 by substituting it back into the
original equation.
32(1)+1 = 27
33= 27
27 = 27
Thus, the solution to the exponential equation 32x+1 = 27 is x= 1.
Question 9
Question
Solve the exponential equation 2x+1 = 8.
Solution
Step 1: Rewrite 8 as a power of 2.
Since 8 = 23,we have 2x+1 = 23.
Step 2: Set the exponents equal to each other.
Therefore, x+ 1 = 3.
Step 3: Solve for x.
Subtracting 1 from both sides gives us x= 2.
Step 4: Verify the solution.
Substitute x= 2 back into the original equation: 22+1 = 23= 8,
which confirms that x= 2 is the correct solution.
Question 10
Question
Samantha invested $5000 in an account earning 3.5% annual interest com-
pounded continuously. How much will she have in the account after 10 years?
7
Solution
Step 1: The formula for compound interest compounded continuously is given
by A=P·ert, where: - Ais the amount of money accumulated after tyears,
-Pis the principal amount (the initial amount of money), - ris the annual
interest rate (written as a decimal), - tis the time the money is invested for (in
years), and - eis the base of the natural logarithm.
Step 2: Given that the principal amount P= $5000, the interest rate r=
0.035 (3.5% written as a decimal), and the time t= 10 years, we can plug these
values into the formula:
A= 5000 ·e0.035·10
Step 3: Simplifying the formula:
A= 5000 ·e0.35
Step 4: Calculating e0.35:
A≈5000 ·1.419067
Step 5: Finding the final amount in the account after 10 years:
A≈7095.335
Therefore, after 10 years, Samantha will have approximately $7095.34 in the
account.
Question 11
Question
Suzanne invests $200 in a savings account that pays 4% interest compounded
quarterly. How much money will be in the account after 5 years?
Solution
Step 1: First, we need to determine the annual interest rate (r) and the number
of compounding periods per year (n). The formula for compound interest is:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after tyears, P= the principal
amount (initial investment), r= annual interest rate (as a decimal), n= number
of compounding periods per year, and t= time in years. Given that the interest
is compounded quarterly, n= 4 and r= 0.04.
Step 2: Let’s substitute the values into the formula and calculate the amount
of money in the account after 5 years:
A= 200 (1 + 0.04
4)4·5
8
A= 200 (1 + 0.01)20
A= 200 ×1.0120
Step 3: Now, we compute 1.0120 to find the final amount in the account
after 5 years.
A≈200 ×1.221386
A≈244.2772
Therefore, after 5 years, Suzanne will have approximately $244.28 in the
savings account.
Question 12
Question
A certain amount of money is invested at an annual interest rate of 5
Solution
Step 1: Given that the initial investment doubles in 10 years, we can express
this information using the continuous compound interest formula:
A=P·ert
where: - Ais the final amount after time t, - Pis the initial amount invested,
-ris the annual interest rate (decimal form), and - tis the time the money is
invested for.
Step 2: Since the initial investment doubles, the final amount after 10 years
is twice the initial amount:
2P=P·e0.05·10
Step 3: Simplify the equation from Step 2 to solve for P:
2 = e0.5
Step 4: Take natural logarithm (ln) of both sides of the equation to eliminate
the exponential:
ln(2) = ln(e0.5)
Step 5: Use the property ln(ex) = xto simplify the equation from Step 4:
ln(2) = 0.5
Step 6: Solve for Pby converting the equation in Step 5 back to exponential
form:
e0.5= 2
Step 7: Therefore, the initial amount of money invested was 2.
9
Question 13
Question
Samantha invests $5000 in a savings account that offers an annual interest rate
of 4% compounded continuously. How much will her investment be worth after
10 years?
Solution
Step 1: Identify the variables and known values in the problem. Let’s denote
the initial amount invested as P= $5000, the annual interest rate as r= 0.04,
the number of years as t= 10, and the amount after tyears as A.
Step 2: Use the formula for continuous compounding to find the future value
of the investment. The formula for continuous compounding is given by:
A=P·ert
Step 3: Substitute the values into the formula and solve for A. Substitute
P= 5000,r= 0.04, and t= 10 into the formula:
A= 5000 ·e0.04·10
Step 4: Calculate the value of A.
A= 5000 ·e0.4
A= 5000 ·e0.4≈5000 ·1.4918 ≈7459.08
Therefore, after 10 years, Samantha’s investment will be worth approxi-
mately $7459.08.
Question 14
Question
Samantha invested some money in a savings account that earns an annual inter-
est rate of 6.5%, compounded quarterly. If she initially invested $10,000, how
much money will she have in the account after 5 years?
Solution
Step 1: Identify the relevant compound interest formula. The formula for com-
pound interest is given by:
A=P(1 + r
n)nt
where: A= the amount of money accumulated after n years, including interest,
P= the principal amount (the initial amount of money), r= annual interest
10
rate (decimal), n= number of times that interest is compounded per year, t=
time the money is invested for in years.
Step 2: Plug in the known values into the formula. In this case, we have:
P= $10,000 (initial investment), r= 6.5% = 0.065 (annual interest rate in
decimal form), n= 4 (interest is compounded quarterly), t= 5 years.
Step 3: Calculate the amount of money Samantha will have in the account
after 5 years using the compound interest formula:
A= 10000 (1 + 0.065
4)4×5
Step 4: Simplify the expression inside the parentheses first:
1 + 0.065
4= 1 + 0.01625 = 1.01625
Step 5: Plug this back into the formula and calculate:
A= 10000 ×(1.01625)20
Step 6: Evaluate (1.01625)20:
(1.01625)20 ≈1.3548
Step 7: Finally, calculate the total amount of money Samantha will have
after 5 years:
A= 10000 ×1.3548 ≈$13,548
Therefore, after 5 years, Samantha will have approximately $13,548 in her
savings account.
Question 15
Question
Solve the exponential equation 32x−6·3x+ 9 = 0.
Solution
Step 1: Let’s make a substitution to simplify the equation. Let y= 3x. Then,
the original equation becomes y2−6y+ 9 = 0.
Step 2: Now, we need to solve the quadratic equation y2−6y+ 9 = 0.
Step 3: Factoring the quadratic equation gives us (y−3)2= 0.
Step 4: To find the value of y, we take the square root of both sides: y−3 = 0.
Step 5: Therefore, y= 3.
Step 6: Now, substitute back 3xfor y:3x= 3.
Step 7: Since 3 = 31, we can write the equation as 3x= 31.
Step 8: Equating the exponents gives x= 1.
Step 9: Thus, the solution to the exponential equation 32x−6·3x+ 9 = 0
is x= 1.
11
Question 16
Question
Solve the exponential equation 23x−1−2x−1= 8 for x.
Solution
Step 1: Rewrite the equation using a common base. Step 2: Use the properties
of exponents to simplify the equation. Step 3: Solve for the variable y. Step 4:
Check the solution in the original equation.
Step 1: Rewrite the equation using a common base.
23x−1−2x−1= 8
Step 2: Use the properties of exponents to simplify the equation.
23x·2−1−2x·2−1= 8
23x·1
2−2x·1
2= 8
23x−1−2x−1= 8
Step 3: Solve for the variable x. Now, the equation becomes:
23x−1−2x−1= 8
Let y= 2x−1, then the equation becomes:
2y3−y= 8
2y3−y−8 = 0
Factoring the equation:
(2y+ 1)(y−2)(y+ 4) = 0
This gives us 3 possible solutions for y:y1=−1
2,y2= 2,y3=−4
Now, we substitute back to solve for x: For y1=−1
2:
2x−1=−1
2
This has no real solution since 2x−1is always positive.
For y2= 2:
2x−1= 2
x−1 = 1
x= 2
12
For y3=−4:
2x−1=−4
This has no real solution since 2x−1is always positive.
Step 4: Check the solution in the original equation. Checking x= 2:
23(2)−1−22−1= 8
25−2 = 8
32 −2 = 8
30 = 8
This solution is not valid, so there is no solution to the equation.
Question 17
Question
Solve the following exponential equation for x:32x+1 = 27.
Solution
Step 1: Rewrite 27 as a power of 3.
Rewrite 27 as 33since 33= 27.
Step 2: Substitute 27 as 33into the equation 32x+1 = 27.
32x+1 = 33.
Step 3: Apply the property of exponents that states if ab=ac, then b=c.
Set 2x+ 1 = 3 to solve for x.
Step 4: Solve for x.
2x+ 1 = 3
2x= 3 −1
2x= 2
x= 1.
Step 5: Check the solution by substituting x= 1 back into the original
equation.
32(1)+1 = 33
33= 27.
Therefore, the solution to the equation 32x+1 = 27 is x= 1.
13
Question 18
Question
A certain radioactive isotope decays at a rate of 8
Solution
Step 1: Let A(t)be the amount of the isotope remaining after thours. Since
the isotope decays at a rate of 8
A(t) = 200 ×(1 −0.08)t
Step 2: Substitute t= 12 into the formula to find the amount of the isotope
remaining after 12 hours:
A(12) = 200 ×(1 −0.08)12
Step 3: Calculate the value of A(12):
A(12) = 200 ×(0.92)12
Step 4: Use a calculator to compute the value of (0.92)12, then multiply the
result by 200 to find the amount of the isotope remaining after 12 hours.
Step 5: After performing the calculations, the amount of the isotope remain-
ing after 12 hours is approximately 102.51 grams.
Question 19
Question
A population of bacteria triples every 5 hours. Initially, there are 100 bacteria
in the population. a) Find the exponential function that models the population
size after thours. b) How many bacteria will be in the population after 20
hours?
Solution
a) Let P(t)be the population size after thours. Since the population triples
every 5 hours, we can write the exponential function as:
P(t) = 100 ·3t/5
b) To find the population after 20 hours, we substitute t= 20 into the
exponential function:
P(20) = 100 ·320/5
P(20) = 100 ·34
P(20) = 100 ·81
P(20) = 8100
After 20 hours, there will be 8100 bacteria in the population.
14
Question 20
Question
Solve the exponential equation 32x−6·3x+ 9 = 0 for x.
Solution
Step 1: Let’s rewrite 32xas (3x)2. This allows us to make a simple substitution.
Step 2: Let y= 3x. Then, the equation becomes y2−6y+ 9 = 0.
Step 3: Factor the quadratic equation y2−6y+ 9 = 0 into (y−3)2= 0.
Step 4: Solve the equation (y−3)2= 0 for y. We get y= 3.
Step 5: Substitute back y= 3xinto 3xto get 3x= 3.
Step 6: Solve the equation 3x= 3 for x. We have x= 1.
Therefore, the solution to the exponential equation 32x−6·3x+ 9 = 0 is
x= 1.
Question 21
Question
Solve the following exponential equation for x:
3x−1= 5.
Solution
Step 1: Rewrite the equation using the properties of exponents:
3x−1= 5
3x= 5 ·31
3x= 15
Step 2: Take the natural logarithm of both sides to solve for x. Remember
the property: log(a·b) = log(a) + log(b).
ln(3x) = ln(15)
xln(3) = ln(15)
x=ln(15)
ln(3)
Step 3: Use a calculator to approximate the value of x:
x≈ln(15)
ln(3) ≈2.7081
1.0986 ≈2.4655
Therefore, the solution to the equation 3x−1= 5 is x≈2.4655.
15
Question 22
Question
A population of bacteria initially has 500 bacteria and doubles every hour. How
many bacteria will there be after 4 hours?
Solution
Step 1: Let P(t)represent the population of bacteria at time tin hours. Since
the population doubles every hour, we have the exponential growth model
P(t) = 500 ·2t.
Step 2: We are interested in finding P(4), which represents the population
after 4 hours. Substitute t= 4 into the model:
P(4) = 500 ·24
Step 3: Compute 24:
24= 16
Step 4: Substitute 24= 16 back into the equation:
P(4) = 500 ·16
Step 5: Calculate the population after 4 hours:
P(4) = 500 ·16 = 8000
Step 6: Therefore, after 4 hours, there will be 8000 bacteria in the popula-
tion.
Question 23
Question
Suppose an investment grows according to the exponential model A(t) = 5000 ·
1.08t, where A(t)represents the amount of money in the account after tyears.
Find the time it takes for the investment to double.
Solution
Step 1: Set up the equation for when the investment doubles. We want to find
twhen A(t) = 2A(0).
5000 ·1.08t= 2 ·5000
1.08t= 2
Step 2: Take the natural logarithm of both sides to solve for t.
ln(1.08t)= ln(2)
16
tln(1.08) = ln(2)
Step 3: Solve for tby dividing by ln(1.08).
t=ln(2)
ln(1.08)
Step 4: Use a calculator to find the approximate value of t.
t≈ln(2)
ln(1.08) ≈0.6931
0.0792 ≈8.75
So, it takes approximately 8.75 years for the investment to double.
Question 24
Question
A certain species of bacteria doubles every 3 hours. If there are 500 bacteria to
start with, how many bacteria will there be after 10 hours? Round your answer
to the nearest whole number.
Solution
Step 1: First, we need to determine the growth rate of the bacteria. Since
the bacteria doubles every 3 hours, the growth rate can be represented by the
function N(t) = 500 ·2t/3, where N(t)is the number of bacteria after thours.
Step 2: We want to find N(10), which represents the number of bacteria
after 10 hours. Plugging in t= 10 into the function, we get:
N(10) = 500 ·210/3
Step 3: Calculating 210/3:
210/3= (21/3)10 = ( 3
√2)10 = 2.154
Step 4: Substitute 2.154 back into N(10):
N(10) = 500 ·2.154
Step 5: Calculate the final answer:
N(10) = 1077
Therefore, there will be approximately 1077 bacteria after 10 hours.
Question 25
Question
Solve the equation log3(x−1) + log3(x+ 3) = 2 for x.
17
Solution
Step 1: Apply the product rule of logarithms to combine the two logarithmic
terms on the left side:
log3((x−1)(x+ 3)) = 2
Step 2: Simplify the equation by expanding the product inside the logarithm:
log3(x2+ 2x−3) = 2
Step 3: Rewrite the logarithmic equation in exponential form:
32=x2+ 2x−3
Step 4: Simplify the exponential equation:
9 = x2+ 2x−3
Step 5: Rearrange the equation into standard quadratic form:
x2+ 2x−12 = 0
Step 6: Factor the quadratic equation:
(x+ 6)(x−2) = 0
Step 7: Set each factor equal to zero and solve for x:
x+ 6 = 0 or x−2 = 0
x=−6or x= 2
Step 8: Check both solutions for extraneous roots by substituting them back
into the original logarithmic equation. Since the logarithm of a negative number
is undefined, x=−6is an extraneous root.
Therefore, the solution to the equation log3(x−1)+log3(x+ 3) = 2 is x= 2.
Question 26
Question
Suppose the population of a city is modeled by the exponential function P(t) =
5000 ·1.02t, where P(t)is the population after tyears. Determine the initial
population and the annual growth rate of the city.
Solution
Step 1: To find the initial population, we evaluate P(0).
P(0) = 5000 ·1.020= 5000 ·1 = 5000
Therefore, the initial population of the city is 5000.
Step 2: The annual growth rate is the constant multiplier in the exponential
function. In this case, the growth rate is 1.02, which means the city’s population
grows by 2% annually.
18
Question 27
Question
A certain species of bacteria doubles in population every 6 hours. If there are
initially 100 bacteria in the population, how many bacteria will there be after
24 hours?
Solution
Step 1: To find the growth factor of the bacteria population, we use the formula
A=P(1 + r)t, where: - Ais the final amount, - Pis the initial amount, - ris
the growth rate, and - tis the time in hours.
Step 2: Since the bacteria doubles every 6 hours, the growth rate is r= 1,
meaning the bacteria population grows by 100
Step 3: Substituting P= 100,r= 1, and t= 24 into the formula, we have:
A= 100(1 + 1)24
Step 4: Simplify the expression inside the parentheses:
A= 100(2)24
Step 5: Calculate 224:
A= 100 ×16,777,216
Step 6: Multiply to find the final amount of bacteria:
A= 1,677,721,600
Therefore, after 24 hours, there will be 1,677,721,600 bacteria in the popu-
lation.
Question 28
Question
Solve the exponential equation for x:3x+2 −3x−1= 20.
Solution
To solve the given exponential equation, we can use the properties of exponents
to manipulate the equation so that we can solve for x.
Step 1: Rewrite the equation using properties of exponents. We
can rewrite the equation as 3x·32−3x·3−1= 20.
Step 2: Simplify the equation. Simplifying the equation gives 9·3x−3x
3=
20.
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Step 3: Combine like terms. Combining like terms yields 8·3x= 20.
Step 4: Solve for x.To solve for x, we first divide both sides by 8:
3x=20
8=5
2.
Step 5: Take the logarithm of both sides. Taking the natural logarithm
of both sides gives ln(3x) = ln (5
2).
Step 6: Apply the power rule of logarithms. Using the power rule, we
have xln(3) = ln (5
2).
Step 7: Solve for x.Finally, we can solve for xby dividing by ln(3):
x=ln(5
2)
ln(3) .
Therefore, the solution to the exponential equation 3x+2 −3x−1= 20 is
x=ln(5
2)
ln(3) .
Question 29
Question
Let f(x) = 2x+ 3 and g(x) = log2(x−1). Find the value(s) of xfor which
f(x) = g(x).
Solution
Step 1: Set f(x)equal to g(x)and solve for x.
2x+ 3 = log2(x−1)
2x+ 3 = ln(x−1)
ln(2) (Using change of base formula)
2x+ 3 = ln(x−1)
ln(2)
2x=ln(x−1)
ln(2) −3
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Step 2: Convert the logarithmic expression to exponential form.
2x=ln(x−1)
ln(2) −3
2x= 2 (1
x−1)−3
2x=2
x−1−3
2x=2−3(x−1)
x−1
2x=2−3x+ 3
x−1
2x=5−3x
x−1
Step 3: Write the equation as a single logarithmic function.
2x=5−3x
x−1
2x=5−3x
x−1
log2(2x) = log2(5−3x
x−1)
x= log2(5−3x
x−1)
Therefore, the value of xfor which f(x) = g(x)is x= log2(5−3x
x−1).
Question 30
Question
A bacteria culture starts with 1000 bacteria and doubles in size every hour.
Write an exponential function to model the population of the bacteria after t
hours. Then, find the population after 5 hours and round to the nearest whole
number.
Solution
Step 1: Let P(t)be the population of the bacteria after thours. Since the
bacteria doubles in size every hour, we can write the exponential function as
P(t) = 1000 ·2t.
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Step 2: To find the population after 5 hours, we substitute t= 5 into the
exponential function:
P(5) = 1000 ·25
P(5) = 1000 ·32
P(5) = 32000
Therefore, the population of the bacteria after 5 hours is 32,000.
Question 31
Question
Solve the exponential equation: 2x−3−2x−2= 12.
Solution
Step 1: Rewrite the equation using a common base. Step 2: Apply properties of
exponents to simplify the equation. Step 3: Solve the resulting linear equation
for 2x. Step 4: Determine the value of xusing the logarithmic function. Step
5: Check the solution for extraneous roots.
Step 1: Rewrite the equation using a common base. We can rewrite the
equation as 2x−3−2x−3·2 = 12.
Step 2: Apply properties of exponents to simplify the equation. Simplifying
further, we have 2x−3(1 −2) = 12, which simplifies to −2·2x−3= 12.
Step 3: Solve the resulting linear equation for 2x. Dividing both sides by
−2gives 2x−3=−6.
Step 4: Determine the value of xusing the logarithmic function. Taking
the logarithm of both sides, we get log(2x−3)= log(−6). Using the property
log(ab)=blog(a)and the fact that the logarithm of a negative number is
undefined, we conclude that the equation has no solution.
Step 5: Check the solution for extraneous roots. Since the logarithm of a
negative number is undefined, there are no real solutions to the original equation.
Question 32
Question
Solve the exponential equation 3x= 27 for x.
Solution
Step 1: Rewrite 27 as a power of 3.
Step 2: 27 = 33, since 27 = 3 ·3·3.
Step 3: Substitute 33back into the original equation and solve for x.
Step 4: 3x= 33
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Step 5: Since the bases are the same, set the exponents equal to each other.
Step 6: x= 3
Therefore the solution to the exponential equation 3x= 27 is x= 3.
Question 33
Question
A population of bacteria starts with 1000 bacteria and triples every 4 hours.
Write an exponential equation representing the population of the bacteria as a
function of time, and determine how many bacteria there will be after 12 hours.
Solution
Step 1: Let P(t)represent the population of bacteria at time t. Since the
population triples every 4 hours, we have exponential growth with a growth
factor of 3. The initial population is 1000 bacteria. Thus, the exponential
equation is given by:
P(t) = 1000 ×3t/4
Step 2: To find the population after 12 hours, substitute t= 12 into the
equation and simplify:
P(12) = 1000 ×312/4= 1000 ×33= 1000 ×27 = 27000
Therefore, after 12 hours, there will be 27,000 bacteria in the population.
Question 34
Question
Solve for x:2x+1 + 2x= 24.
Solution
Step 1: Simplify the left side of the equation by factoring out a common factor.
2x+1 + 2x= 2 ·2x+ 2x
= 2 ·2x+ 1 ·2x
= (2 + 1) ·2x
= 3 ·2x
23
Step 2: Substitute the simplified expression back into the equation and solve
for x.
3·2x= 24
2x=24
3
2x= 8
Step 3: Rewrite 8 as a power of 2 to help solve for x.
2x= 23
Step 4: Since the bases are the same, the exponents must be equal.
x= 3
Therefore, the solution to the equation 2x+1 + 2x= 24 is x= 3.
Question 35
Question
Solve the exponential equation 3x+1
27 = 9.
Solution
Step 1: Rewrite both sides of the equation with the same base:
3x+1
27 = 32
Step 2: Rewrite 27 as 33:
3x+1
33= 32
Step 3: Apply the quotient rule of exponents to simplify the left side:
3x+1−3= 32
Step 4: Simplify the exponent on the left side:
3x−2= 32
Step 5: Since the bases are the same, set the exponents equal to each other:
x−2 = 2
Step 6: Solve for x:
x= 2 + 2
Step 7: Simplify to find the final solution:
x= 4
Therefore, the solution to the exponential equation 3x+1
27 = 9 is x= 4.
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