MATH 121 - COLLEGE ALGEBRA Applications of exponential and logarithmic functions Question Bank Set 1

MATH 121 - COLLEGE ALGEBRA -

Applications of exponential and

logarithmic functions

Question Bank - Set 1

Liberty University

Question 1

Question

Samantha invested $5000 in a savings account that earns 3.5% annual interest.

How long will it take for her investment to double?

Solution

Step 1: Let’s denote the amount of time it will take for Samantha’s investment

to double as tyears. Since the investment is doubling, this means that the final

amount is twice the initial amount, so the final amount will be 2×$5000 =

$10000.

Step 2: We can use the formula for compound interest to calculate the final

amount:

A=P(1 + r

n)nt

where: - Ais the final amount ($10000), - Pis the principal amount ($5000), - r

is the annual interest rate (3.5% = 0.035), - nis the number of times the interest

is compounded per year (assume compound interest is compounded annually),

-tis the number of years.

Step 3: Substituting the known values into the formula, we get:

10000 = 5000 (1 + 0.035

1)1·t

Step 4: Simplifying the equation, we have:

2 = (1.035)t

Step 5: To solve for t, we need to take the natural logarithm of both sides:

ln(2) = ln (1.035t)

ln(2) = t·ln(1.035)

Step 6: Finally, we solve for tby dividing both sides by ln(1.035):

t=ln(2)

ln(1.035) ≈0.6931

0.034

]≈20.3859 years

Step 7: Therefore, it will take approximately 20.39 years for Samantha’s

investment to double.

Question 2

Question

Solve the following exponential equation for x:32x= 27.

Solution

Step 1: Rewrite 27 as a power of 3, so 27 = 33.

Step 2: Substitute 33for 27 in the original equation to get 32x= 33.

Step 3: By the property of exponents that states when two powers with the

same base are equal, their exponents must also be equal, we have 2x= 3.

Step 4: Solve for xby dividing both sides of the equation by 2: x=3

Therefore, the solution to the equation 32x= 27 is x=3

Question 3

Question

Solve the exponential equation 3x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3.

Since 27 = 33,the equation becomes 3x−1= 33.

Step 2: Set the exponents equal to each other.

Since 3x−1= 33,we have x−1 = 3.

Step 3: Solve for x.

Adding 1 to both sides gives x= 4.

Therefore, the solution to the exponential equation 3x−1= 27 is x= 4.

Question 4

Question

Samantha invested $10,000 in a savings account that earns 3.5

Solution

Let’s denote the amount of money in the savings account after time tin years

as A(t). Since the interest is compounded continuously, we can use the formula

A(t) = P·ert, where Pis the principal amount ($10,000 in this case), ris the

annual interest rate (0.035), and tis the time in years.

To determine how long it will take for the initial investment to double, we

need to find the value of twhen A(t) = 2 ·P.

Step 1: Start by substituting the given values into the formula.

A(t) = 10000 ·e0.035t

Step 2: Set up the equation to find when the initial investment will double.

2·10000 = 10000 ·e0.035t

Step 3: Simplify the equation.

20000 = 10000 ·e0.035t

Step 4: Divide both sides by 10000 to isolate the exponential term.

2 = e0.035t

Step 5: Take the natural logarithm of both sides to solve for t.

ln(2) = ln(e0.035t)

Step 6: Use the property ln(ex) = x.

ln(2) = 0.035t

Step 7: Solve for t.

t=ln(2)

0.035 ≈19.86 years

Therefore, it will take approximately 19.86 years for Samantha’s initial in-

vestment to double in a savings account with 3.5

Question 5

Question

Solve the exponential equation 32x−1= 27 for x.

Solution

Step 1: Rewrite 27 as a power of 3, which is 33.

Step 2: Substitute 33back into the equation and solve for x.

32x−1= 33

2x−1 = 3 (Since the bases are the same, we can set the exponents equal)

2x= 4

x= 2

Step 3: Check the solution by substituting x= 2 back into the original

equation.

32(2)−1= 27

33= 27

27 = 27 (True)

Therefore, the solution to the equation 32x−1= 27 is x= 2.

Question 6

Question

A certain investment doubles every 5 years. If the initial investment was P,

express the amount of the investment as a function of time tin years. Also,

determine how long it will take for the investment to triple in value if the initial

investment was 100,000.

Solution

Let A(t)be the amount of the investment at time tyears. Since the investment

doubles every 5 years, we have A(t) = P·2t

Step 1: Express the amount of the investment as a function of time in

years. Given that the initial investment was P, we have A(0) = P. Therefore,

the amount of the investment as a function of time tis A(t) = P·2t

Step 2: Determine how long it will take for the investment to triple in value.

If the initial investment was 100,000, wehaveP = 100,000.W eneedtof indtsuchthatA(t)

= 3 ·100,000.

3·100,000 = 100,000 ·2t

300,000 = 2 t

5·100,000

3 = 2 t

To solve for t, take the logarithm of both sides:

log 3 = log 2 t

log 3 = t

5log 2

t= 5log 3

log 2

t≈7.977 years

Therefore, it will take approximately 7.977 years for the investment to triple in

value if the initial investment was 100,000.

Question 7

Question

Solve the equation 3x−2−27 = 0.

Solution

Step 1: Add 27 to both sides of the equation to isolate 3x−2.

3x−2= 27

Step 2: Rewrite 27 as 33since 33= 27.

3x−2= 33

Step 3: Since the bases are the same, set the exponents equal to each other.

x−2 = 3

Step 4: Add 2 to both sides of the equation to solve for x.

x= 3 + 2

Step 5: Simplify the expression.

x= 5

Therefore, the solution to the equation 3x−2−27 = 0 is x= 5.

Question 8

Question

Samantha invests $4000 in a savings account with an annual interest rate of

3.5% compounded continuously.

1. Find a formula for the balance in the account after tyears.

2. How long will it take for Samantha’s investment to double?

Solution

Let A(t)represent the balance in the account after tyears.

1. Step 1: We can use the formula for continuously compounded interest,

which is given by:

A(t) = P·ert,

where: P= $4000 (initial investment), r= 0.035 (annual interest rate in

decimal form), and tis the time in years.

Substituting the values, we get:

A(t) = 4000 ·e0.035t.

2. Step 2: To find when the investment will double, we need to solve for t

in the equation:

4000 ·e0.035t= 8000.

Step 3: Divide both sides by 4000:

e0.035t= 2.

Step 4: Take the natural logarithm of both sides to solve for t:

ln(e0.035t)= ln(2).

0.035t= ln(2).

Step 5: Solve for t:

t=ln(2)

0.035 ≈19.86 years.

So, Samantha’s investment will double in approximately 19.86 years.

Question 9

Question

The population of a city is currently 500,000 and is projected to increase by 2

Solution

Step 1: Let’s denote the initial population of the city as P0= 500,000, the

annual growth rate as r= 2% = 0.02, and the number of years as t= 10.

Step 2: The exponential growth model for the population of the city can be

expressed as:

P(t) = P0·(1 + r)t

Step 3: Substitute the given values into the model:

P(10) = 500,000 ·(1 + 0.02)10

Step 4: Simplify the expression:

P(10) = 500,000 ·(1.02)10

Step 5: Calculate (1.02)10:

(1.02)10 = 1.218994276

Step 6: Substitute the value back into the expression and calculate the

population after 10 years:

P(10) = 500,000 ·1.218994276

P(10) ≈609,497

Therefore, the population of the city is projected to be approximately 609,497

after 10 years.

Question 10

Question

Solve for x:22x+1 −2x+3 = 8.

Solution

Step 1: Rewrite the equation using a common base.

We know that 8 = 23. By expressing 8in terms of 2, we can rewrite the equation

as:

22x+1 −2x+3 = 23

Step 2: Simplify the equation using exponent properties.

Using the properties of exponents, we can rewrite the equation as:

22x·2−2x·23= 23

Step 3: Apply distributive property of exponents.

Simplify the equation further by distributing the exponents:

22x·2−2x+3 = 23

Step 4: Rewrite the equation using simpler terms.

Substitute 2x+3 = 2x·23back into the equation to simplify it:

22x·2−2x·23= 23

Step 5: Combine like terms.

Combine like terms on the left side of the equation:

22x·2−8·2x= 23

Step 6: Factor out common terms.

Factor out a 2xfrom the left side of the equation:

2x·(22−8) = 23

Step 7: Simplify the terms.

Calculate 22−8:

2x·4−8 = 23

Step 8: Solve for x.

Solve the equation:

4·2x−8 = 8

4·2x= 16

2x= 4

x= 2

Step 9: Verify the solution.

Substitute x= 2 back into the original equation to verify the solution:

22(2)+1 −22+3 = 8

25−25= 8

32 −32 = 8

0 = 8

Since the final equation is false, there is no solution to the original equation.

Question 11

Question

Solve for xin the equation 3·4x−1−2 = 25.

Solution

Step 1: Start by isolating the exponential term by adding 2 to both sides of the

equation:

3·4x−1= 25 + 2

Step 2: Simplify the right side of the equation:

3·4x−1= 27

Step 3: Divide both sides by 3 to isolate 4x−1:

4x−1= 9

Step 4: Rewrite the equation using the equivalent exponential form:

x−1 = log4(9)

Step 5: Use the definition of a logarithm to rewrite the equation as an

exponential form:

4x−1= 9

Step 6: Since 9 = 32, rewrite the equation using the same base:

4x−1= 42

Step 7: Set the exponents equal to each other:

x−1 = 2

Step 8: Add 1 to both sides to solve for x:

x= 3

Therefore, the solution to the equation is x= 3.

Question 12

Question

Solve the equation 2x= 5 −3xfor x.

Solution

Step 1: Rewrite the equation as 2x+ 3x= 5.

Step 2: Since 2xand 3xare both positive for all x, it follows that 2x+ 3x>0

for all x.

Step 3: Observe that x= 2 is a solution to the equation since 22+ 32=

4 + 9 = 13 = 5.

Step 4: To prove that x= 2 is the only solution, we will calculate the

derivatives of both sides of the equation.

Step 5: Let f(x) = 2x+ 3x, and find f′(x).

Step 6: f′(x) = ln(2) ·2x+ ln(3) ·3x.

Step 7: Let g(x) = 5, and find g′(x).

Step 8: g′(x) = 0.

Step 9: Comparing f′(x)and g′(x), it is clear that f′(x)=g′(x)for all x,

which means x= 2 is the only solution to the equation 2x+ 3x= 5.

Therefore, the solution to the equation 2x= 5 −3xis x= 2.

Question 13

Question

Solve the exponential equation 3x−2= 81 for x.

Solution

Step 1: Rewrite 81 as a power of 3.

81 = 34

Step 2: Substitute 34for 81 in the original equation and simplify.

3x−2= 34

x−2 = 4

Step 3: Solve for xby adding 2 to both sides of the equation.

x= 4 + 2

x= 6

Step 4: Check the solution by substituting x= 6 into the original equation.

36−2= 81

34= 81

81 = 81

Therefore, the solution to the equation 3x−2= 81 is x= 6.

Question 14

Question

Solve for xin the equation 23x= 5.

Solution

Step 1: Take the natural logarithm of both sides to eliminate the exponential

function.

ln(23x)= ln(5)

Step 2: Apply the logarithmic property ln(ab)=bln(a)to simplify the left

side.

3xln(2) = ln(5)

Step 3: Divide both sides by 3ln(2) to solve for x.

x=ln(5)

3 ln(2)

Step 4: Use the property logb(a) = ln(a)

ln(b)to rewrite xin a more simplified

form.

x=ln(5)

3 ln(2) =loge(5)

3 loge(2)

So, the solution to the equation 23x= 5 is x=loge(5)

3 loge(2) .

Question 15

Question

Samantha invested $5000 in a high yield savings account with an annual interest

rate of 4.5

Solution

Step 1: First, we identify the formula for continuous compound interest:

A=P·ert

where: - Ais the amount of money accumulated after tyears, - Pis the principal

amount (initial investment), - ris the annual interest rate (in decimal form),

and - tis the time the money is invested for.

Step 2: Plug in the given values into the formula:

A= 5000 ·e0.045×10

Step 3: Calculate the amount after 10 years:

A= 5000 ·e0.45

Step 4: Use the approximation e≈2.71828 to evaluate the expression:

A≈5000 ·2.718280.45

Step 5: Compute the final result:

A≈5000 ·1.56934 ≈7846.7

Therefore, Samantha will have approximately $7846.7 in her account after

10 years.

Question 16

Question

Solve the exponential equation for x:52x+1 = 125.

Solution

Step 1: Rewrite 125 as a power of 5.

125 = 53

Step 2: Set the exponents equal to each other and solve for x.

2x+ 1 = 3

2x= 2

x= 1

Step 3: Check the solution by substituting x= 1 back into the original

equation.

52(1)+1 = 53

53= 125

Therefore, the solution to the exponential equation is x= 1.

Question 17

Question

Let f(x) = log2(x2−1) be a logarithmic function. Find the domain of the

function f(x).

Solution

Step 1: The domain of a logarithmic function f(x) = logb(u(x)) is the set of

all real numbers xsuch that u(x)>0. Here, u(x) = x2−1and the base of the

logarithm is b= 2.

Step 2: Set the expression inside the logarithm greater than zero and solve

for x:

x2−1>0

Step 3: Factor the quadratic inequality:

(x−1)(x+ 1) >0

Step 4: The critical points are x=−1and x= 1. We create a sign chart

to analyze where the inequality is true:

x < −1−1< x < 1x > 1

x−1− − +

x+ 1 −+ +

(x−1)(x+ 1) + −+

Step 5: The inequality (x−1)(x+ 1) >0is true when x < −1or x > 1.

Therefore, the domain of f(x)is (−∞,−1) ∪(1,∞).

Question 18

Question

Solve the exponential equation 32x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3:

27 = 33

Step 2: Substitute 33for 27 in the equation:

32x−1= 33

Step 3: Since the bases are the same, we can set the exponents equal to each

other:

2x−1 = 3

Step 4: Solve for x:

2x−1 = 3

2x= 4

x= 2

Step 5: Check the solution by substituting x= 2 back into the original

equation:

32(2)−1= 33

33= 33

Since the equation holds true, the solution x= 2 is correct.

Question 19

Question

Solve the exponential equation 32x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3.

27 = 33

Step 2: Set the exponential equation in terms of the base 3.

32x−1= 33

Step 3: Equate the exponents.

2x−1 = 3

Step 4: Solve for x.

2x= 4

x= 2

Step 5: Check the solution. Substitute x= 2 back into the original equation:

32(2)−1= 33

33= 27

Since the left side equals the right side, x= 2 is the correct solution to the

exponential equation 32x−1= 27.

Question 20

Question

A certain radioactive substance decays over time. The amount of the substance

present after tyears is given by the function A(t) = A0e−kt, where A0is the

initial amount of the substance and kis a positive constant. If 30

Solution

Step 1: Given that 30

A(4) = A0e−4k= 0.7A0

Step 2: Divide both sides of the equation by A0to simplify the equation.

e−4k= 0.7

Step 3: Take the natural logarithm of both sides to solve for k.

ln(e−4k)= ln(0.7)

−4k= ln(0.7)

Step 4: Solve for kby dividing both sides by −4.

k=ln(0.7)

−4=−ln(7)

Therefore, the value of kis −ln(7)

Question 21

Question

Solve the exponential equation 23x+1 −2x+2 = 8.

Solution

Step 1: We can rewrite 8as 23. Therefore, the equation becomes 23x+1 −2x+2 =

23.

Step 2: By using the properties of exponents, we can rewrite 23x+1 as 2·23x

and 2x+2 as 2·2x. Substituting these into the equation gives us 2·23x−2·2x= 23.

Step 3: Factor out a 2xterm from both terms on the left side, giving us

2x(2 ·22x−2) = 23.

Step 4: Simplify the expression inside the parentheses to get 2x(4 ·2x−2) =

23.

Step 5: Further simplifying, we have 2x(2x+2 −2) = 23.

Step 6: Applying the property of exponents again, we find that 2x+2 = 4·2x.

Substituting this back into the equation gives us 2x(4 ·2x−2) = 23.

Step 7: Solve for 2xby dividing both sides by (4 ·2x−2):2x=23

4·2x−2.

Step 8: Substitute 2x=yto simplify the equation to y=8

4y−2.

Step 9: Multiply both sides by (4y−2) to get y(4y−2) = 8.

Step 10: Expand and rearrange the equation to get 4y2−2y−8 = 0.

Step 11: Solve this quadratic equation by factoring or using the quadratic

formula to find the value(s) of y.

Step 12: Once you have found the value(s) of y, substitute back to find the

corresponding values of 2x. Remember to check for extraneous solutions.

Question 22

Question

A certain population of bacteria doubles every 3 hours. If there are 100 initial

bacteria, how many bacteria will there be after 12 hours?

Solution

Step 1: Determine the rate of growth for the bacteria population.

Given that the population doubles every 3 hours, we can express this ex-

ponential growth using the formula for exponential growth: A(t) = A0·(2) t

where: - A(t)is the population after thours, - A0is the initial population, - 2

is the factor by which the population doubles every 3 hours, and - t

3represents

the number of 3-hour intervals.

Step 2: Substitute the given values into the formula.

Given that the initial population is 100 and we want to find the population

after 12 hours, we have: A(12) = 100 ·(2)12

Step 3: Solve for the final population after 12 hours.

A(12) = 100 ·24

A(12) = 100 ·16

A(12) = 1600

After 12 hours, there will be 1600 bacteria in the population.

Question 23

Question

A population of bacteria triples every 5 minutes in a lab experiment. If there

were initially 100 bacteria, how long will it take for the population to reach 5000

bacteria?

Solution

Step 1: Let’s denote the initial number of bacteria as P0and the time it takes

for the population to reach a certain number of bacteria as t. Given that the

population triples every 5 minutes, we can express the population growth as an

exponential function: P(t) = P0·3t/5.

Step 2: We are given that P0= 100 bacteria. Substituting this value into

the exponential function, we have P(t) = 100 ·3t/5.

Step 3: We also know that we want to find the time twhen the population

reaches 5000 bacteria. So, we have the equation 5000 = 100 ·3t/5.

Step 4: Dividing both sides by 100, we get 50 = 3t/5.

Step 5: Taking the natural logarithm of both sides, we have ln(50) =

ln(3t/5).

Step 6: Using the properties of logarithms, we can rewrite the right side as

t/5·ln(3).

Step 7: Simplifying, we find t= 5 ·ln(50)

ln(3) .

Step 8: Using a calculator to approximate, we find t≈24.57 minutes. There-

fore, it will take approximately 24.57 minutes for the population to reach 5000

bacteria.

Question 24

Question

Solve the exponential equation: 32x−1= 9.

Solution

Step 1: Rewrite both sides of the equation with the same base.

32x−1= 9

32x−1= 32

Step 2: Since the bases are equal, we can set the exponents equal to one

another.

2x−1 = 2

Step 3: Solve for x.

2x−1 = 2

2x= 3

x=3

x= 1.5

Step 4: Therefore, the solution to the exponential equation 32x−1= 9 is

x= 1.5.

Question 25

Question

Solve the exponential equation 32x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3:

27 = 33

Step 2: Substitute 33for 27 in the equation:

32x−1= 33

Step 3: Since the bases are the same, set the exponents equal to each other:

2x−1 = 3

Step 4: Solve for xby isolating x:

2x= 4

x= 2

Step 5: Check the solution by substituting x= 2 back into the original

equation:

32(2)−1= 33

33= 33

Step 6: Since the left side equals the right side, x= 2 is the correct solution.

Question 26

Question

Suppose a radioactive substance decays according to the formula A(t) = A0e−kt,

where A(t)is the amount of substance remaining after tdays, A0is the initial

amount of substance, and kis a constant. If 100 grams of a substance decay to

70 grams in 10 days, find the value of k.

Solution

Step 1: We are given that A0= 100 grams, A(10) = 70 grams, and t= 10 days.

Substituting these values into the formula A(t) = A0e−kt, we get:

70 = 100e−10k

Step 2: Divide both sides by 100 to isolate the exponential term:

100 =e−10k

Step 3: Simplify the left side:

0.7 = e−10k

Step 4: Take the natural logarithm of both sides to solve for k:

ln(0.7) = ln(e−10k)

ln(0.7) = −10kln(e)

ln(0.7) = −10k

Step 5: Solve for kby dividing both sides by −10:

k=ln(0.7)

−10

Therefore, the value of kis k=ln(0.7)

−10 .

Question 27

Question

Solve the exponential equation 32x−1−3x= 20 for x.

Solution

Step 1: Let’s rewrite the equation in a way that makes it easier to solve by using

an intermediate variable. Step 2: Let y= 3x. Step 3: Substituting y= 3xinto

the equation gives us 32x−1−y= 20. Step 4: Now, we have 3·32x−1−y= 20.

Step 5: Simplifying the left side of the equation gives us 32x−y= 20. Step

6: Since y= 3x, we can rewrite the equation as 32x−3x= 20. Step 7: To

solve this equation, let’s rewrite 32xas (3x)2so that it resembles a quadratic

equation. Step 8: Let u= 3x. Now, our equation becomes u2−u= 20. Step

9: Rearranging terms gives us the quadratic equation u2−u−20 = 0. Step

10: Factoring the quadratic equation gives us (u−5)(u+ 4) = 0. Step 11:

Setting each factor to zero gives us u−5=0or u+ 4 = 0. Step 12: Solving

these equations, we find u= 5 or u=−4. Step 13: Since u= 3x, we have two

possible solutions: 3x= 5 or 3x=−4. Step 14: The equation 3x=−4has

no real solutions, so we focus on solving 3x= 5. Step 15: Taking the natural

logarithm of both sides, we get ln(3x) = ln(5). Step 16: Applying the logarithm

property ln(ab)=bln(a)gives us xln(3) = ln(5). Step 17: Therefore, the

solution to the equation is x=ln(5)

ln(3) .

Question 28

Question

Solve the following exponential equation for x:

32x+1 = 27

Solution

Step 1: Rewrite 27 as a power of 3.

27 = 33

Step 2: Therefore, the equation becomes:

32x+1 = 33

Step 3: Since the bases are the same, we can equate the exponents:

2x+ 1 = 3

Step 4: Subtract 1 from both sides to isolate 2x:

2x= 2

Step 5: Finally, divide by 2 to solve for x:

x= 1

Therefore, the solution to the given exponential equation is x= 1.

Question 29

Question

The population of a city is currently 50,000 and is growing at a rate of 2

Solution

Step 1: We can model the population growth using the exponential growth

formula:

P(t) = P0·(1 + r)t

where: - P(t)is the population after tyears, - P0is the initial population, - r

is the growth rate per year written as a decimal, and - tis the time in years.

Step 2: In this case, we are given that P0= 50,000,r= 0.02, and we want

to find the time it takes for the population to reach 75,000. So, we set up the

equation:

75,000 = 50,000 ·(1 + 0.02)t

Step 3: Next, we solve for t. Divide both sides by 50,000 to isolate the

exponential term: 75,000

50,000 = (1.02)t

Step 4: Simplify the left side:

1.5 = (1.02)t

Step 5: Now, to solve for t, take the natural logarithm of both sides:

ln(1.5) = ln(1.02)t

Step 6: By properties of logarithms, we can bring the exponent down in

front:

ln(1.5) = t·ln(1.02)

Step 7: Divide both sides by ln(1.02) to solve for t:

t=ln(1.5)

ln(1.02)

Step 8: Calculate the value of tusing a calculator:

t≈ln(1.5)

ln(1.02) ≈0.4055

0.0198 ≈20.48

Step 9: It will take approximately 20.48 years for the population to reach

75,000.

Question 30

Question

Solve the exponential equation: 3x−2= 7.

Solution

Step 1: Rewrite the equation in exponential form to get rid of the exponent.

3x−2= 7

3x−2= 3log37

Step 2: Since the bases are the same, set the exponents equal to each other.

x−2 = log37

Step 3: Solve for xby adding 2 to both sides of the equation.

x= log37+2

Step 4: Use the change of base formula to rewrite the logarithmic expression.

x=log 7

log 3 + 2

Therefore, the solution to the exponential equation 3x−2= 7 is x=log 7

log 3 + 2.

Question 31

Question

Samantha invests $4000 in a savings account that earns 4% annual interest

compounded quarterly. How much will Samantha have in her account after 5

years?

Solution

Step 1: First, we need to determine the annual interest rate when compounded

quarterly. Since the interest is compounded quarterly, the quarterly interest

rate ris given by:

r=4%

4= 1% = 0.01

Step 2: Next, we calculate the total number of compounding periods over

5 years. Since the interest is compounded quarterly, the total number of com-

pounding periods nis given by:

n= 4 ×5 = 20

Step 3: We can now use the formula for compound interest to find the final

amount in Samantha’s account. The formula is given by:

A=P(1 + r

n)nt

where: - Ais the amount of money accumulated after tyears. - Pis the

principal amount (initial investment) ($4000 in this case). - ris the interest

rate per compounding period (quarterly rate of 0.01 in this case). - nis the

number of compounding periods per year (4 in this case). - tis the time the

money is invested for (5 years in this case).

Step 4: Plugging in the values, we have:

A= 4000 (1 + 0.01

4)4×5

Step 5: Calculate the amount Samantha will have in her account after 5

years.

A= 4000 (1.0025)20

A= 4000 ×1.220218 . . .

A≈$4880.87

Therefore, Samantha will have approximately $4880.87 in her account after

5 years.

Question 32

Question

Samantha invests $5000 in an account that pays an annual interest rate of 4%.

Assuming the interest is compounded continuously, how much money will be in

the account after 10 years?

Solution

Step 1: Let’s denote the amount of money in the account after tyears as A(t).

The formula for continuously compounded interest is given by A(t) = P·ert,

where Pis the principal amount, ris the interest rate (in decimal form), tis the

time in years, and eis the base of the natural logarithm. In this case, P= 5000,

r= 0.04, and t= 10.

Step 2: Substituting the given values into the formula, we get:

A(10) = 5000 ·e0.04·10

Step 3: Simplifying the expression gives:

A(10) = 5000 ·e0.4

Step 4: Using the approximation e≈2.71828, we can further simplify:

A(10) ≈5000 ·2.718280.4

Step 5: Calculating the value of A(10), we find:

A(10) ≈5000 ·1.491824

Step 6: Therefore, the amount of money in the account after 10 years,

rounded to the nearest dollar, is approximately $7459.

Question 33

Question

Solve the following exponential equation for x:23x−1= 64.

Solution

Step 1: Rewrite 64 as a power of 2.

Since 64 = 26,we have 23x−1= 26.

Step 2: Set the exponents equal to each other.

This gives us the equation 3x−1 = 6.

Step 3: Solve for x.

3x−1 = 6

3x= 7

x=7

Step 4: Check the solution.

Plug x=7

3back into the original equation: 23( 7

3)−1= 26−1= 25= 32 = 64.

Since the solution does not satisfy the original equation, there is no solution

to the equation 23x−1= 64.

Question 34

Question

Sara invested $5000 in a savings account that pays 3.5% annual interest, com-

pounded quarterly. How much money will be in the account after 5 years?

Solution

Step 1: Identify the variables given in the problem. Let’s denote: - P= $5000

(the principal amount invested), - r= 0.035 (the annual interest rate as a

decimal), - n= 4 (the number of times the interest is compounded per year), -

t= 5 (the number of years the money is invested for).

Step 2: Use the compound interest formula to find the future value of the

investment after 5 years. The compound interest formula is given by:

A=P(1 + r

n)nt

Plugging in the values we have:

A= 5000 (1 + 0.035

4)4×5

Step 3: Simplify the equation by calculating inside the parentheses first.

A= 5000 (1 + 0.00875)20

Step 4: Calculate the value inside the parentheses.

A= 5000(1.00875)20

Step 5: Raise 1.00875 to the 20th power.

A≈5000 ×1.2059

Step 6: Calculate the final answer.

A≈6029.50

Therefore, after 5 years, there will be approximately $6029.50 in the account.

Question 35

Question

A population of bacteria doubles every 3 hours. Initially, there are 100 bacteria

in the population. Write a formula for the number of bacteria after thours.

Solution

Step 1: Let N(t)represent the number of bacteria after thours. Since the

population doubles every 3 hours, we can express this exponential growth using

the formula:

N(t) = 100 ·2t

Step 2: Substitute t= 0 into the formula to find the initial number of

bacteria:

N(0) = 100 ·20

3= 100 ·20= 100

Therefore, with an initial population of 100 bacteria, the formula for the

number of bacteria after thours is N(t) = 100 ·2t

Question 4

Question

Samantha invested $10,000 in a savings account that earns 3.5

Solution

Let’s denote the amount of money in the savings account after time tin years

as A(t). Since the interest is compounded continuously, we can use the formula

A(t) = P·ert, where Pis the principal amount ($10,000 in this case), ris the

annual interest rate (0.035), and tis the time in years.

To determine how long it will take for the initial investment to double, we

need to find the value of twhen A(t) = 2 ·P.

Step 1: Start by substituting the given values into the formula.

A(t) = 10000 ·e0.035t

Step 2: Set up the equation to find when the initial investment will double.

2·10000 = 10000 ·e0.035t

Step 3: Simplify the equation.

20000 = 10000 ·e0.035t

Step 4: Divide both sides by 10000 to isolate the exponential term.

2 = e0.035t

Step 5: Take the natural logarithm of both sides to solve for t.

ln(2) = ln(e0.035t)

Step 6: Use the property ln(ex) = x.

ln(2) = 0.035t

Step 7: Solve for t.

t=ln(2)

0.035 ≈19.86 years

Therefore, it will take approximately 19.86 years for Samantha’s initial in-

vestment to double in a savings account with 3.5

Question 5

Question

Solve the exponential equation 32x−1= 27 for x.

Solution

Step 1: Rewrite 27 as a power of 3, which is 33.

Step 2: Substitute 33back into the equation and solve for x.

32x−1= 33

2x−1 = 3 (Since the bases are the same, we can set the exponents equal)

2x= 4

x= 2

Step 3: Check the solution by substituting x= 2 back into the original

equation.

32(2)−1= 27

33= 27

27 = 27 (True)

Therefore, the solution to the equation 32x−1= 27 is x= 2.

Question 6

Question

A certain investment doubles every 5 years. If the initial investment was P,

express the amount of the investment as a function of time tin years. Also,

determine how long it will take for the investment to triple in value if the initial

investment was 100,000.

Solution

Let A(t)be the amount of the investment at time tyears. Since the investment

doubles every 5 years, we have A(t) = P·2t

Step 1: Express the amount of the investment as a function of time in

years. Given that the initial investment was P, we have A(0) = P. Therefore,

the amount of the investment as a function of time tis A(t) = P·2t

Step 2: Determine how long it will take for the investment to triple in value.

If the initial investment was 100,000, wehaveP = 100,000.W eneedtof indtsuchthatA(t)

= 3 ·100,000.

3·100,000 = 100,000 ·2t

300,000 = 2 t

5·100,000

3 = 2 t

To solve for t, take the logarithm of both sides:

log 3 = log 2 t

log 3 = t

5log 2

t= 5log 3

log 2

t≈7.977 years

Therefore, it will take approximately 7.977 years for the investment to triple in

value if the initial investment was 100,000.

Question 7

Question

Solve the equation 3x−2−27 = 0.

Solution

Step 1: Add 27 to both sides of the equation to isolate 3x−2.

3x−2= 27

Step 2: Rewrite 27 as 33since 33= 27.

3x−2= 33

Step 3: Since the bases are the same, set the exponents equal to each other.

x−2 = 3

Step 4: Add 2 to both sides of the equation to solve for x.

x= 3 + 2

Step 5: Simplify the expression.

x= 5

Therefore, the solution to the equation 3x−2−27 = 0 is x= 5.

Question 8

Question

Samantha invests $4000 in a savings account with an annual interest rate of

3.5% compounded continuously.

1. Find a formula for the balance in the account after tyears.

2. How long will it take for Samantha’s investment to double?

Solution

Let A(t)represent the balance in the account after tyears.

1. Step 1: We can use the formula for continuously compounded interest,

which is given by:

A(t) = P·ert,

where: P= $4000 (initial investment), r= 0.035 (annual interest rate in

decimal form), and tis the time in years.

Substituting the values, we get:

A(t) = 4000 ·e0.035t.

2. Step 2: To find when the investment will double, we need to solve for t

in the equation:

4000 ·e0.035t= 8000.

Step 3: Divide both sides by 4000:

e0.035t= 2.

Step 4: Take the natural logarithm of both sides to solve for t:

ln(e0.035t)= ln(2).

0.035t= ln(2).

Step 5: Solve for t:

t=ln(2)

0.035 ≈19.86 years.

So, Samantha’s investment will double in approximately 19.86 years.

Question 9

Question

The population of a city is currently 500,000 and is projected to increase by 2

Solution

Step 1: Let’s denote the initial population of the city as P0= 500,000, the

annual growth rate as r= 2% = 0.02, and the number of years as t= 10.

Step 2: The exponential growth model for the population of the city can be

expressed as:

P(t) = P0·(1 + r)t

Step 3: Substitute the given values into the model:

P(10) = 500,000 ·(1 + 0.02)10

Step 4: Simplify the expression:

P(10) = 500,000 ·(1.02)10

Step 5: Calculate (1.02)10:

(1.02)10 = 1.218994276

Step 6: Substitute the value back into the expression and calculate the

population after 10 years:

P(10) = 500,000 ·1.218994276

P(10) ≈609,497

Therefore, the population of the city is projected to be approximately 609,497

after 10 years.

Question 10

Question

Solve for x:22x+1 −2x+3 = 8.

Solution

Step 1: Rewrite the equation using a common base.

We know that 8 = 23. By expressing 8in terms of 2, we can rewrite the equation

as:

22x+1 −2x+3 = 23

Step 2: Simplify the equation using exponent properties.

Using the properties of exponents, we can rewrite the equation as:

22x·2−2x·23= 23

Step 3: Apply distributive property of exponents.

Simplify the equation further by distributing the exponents:

22x·2−2x+3 = 23

Step 4: Rewrite the equation using simpler terms.

Substitute 2x+3 = 2x·23back into the equation to simplify it:

22x·2−2x·23= 23

Step 5: Combine like terms.

Combine like terms on the left side of the equation:

22x·2−8·2x= 23

Step 6: Factor out common terms.

Factor out a 2xfrom the left side of the equation:

2x·(22−8) = 23

Step 7: Simplify the terms.

Calculate 22−8:

2x·4−8 = 23

Step 8: Solve for x.

Solve the equation:

4·2x−8 = 8

4·2x= 16

2x= 4

x= 2

Step 9: Verify the solution.

Substitute x= 2 back into the original equation to verify the solution:

22(2)+1 −22+3 = 8

25−25= 8

32 −32 = 8

0 = 8

Since the final equation is false, there is no solution to the original equation.

Question 11

Question

Solve for xin the equation 3·4x−1−2 = 25.

Solution

Step 1: Start by isolating the exponential term by adding 2 to both sides of the

equation:

3·4x−1= 25 + 2

Step 2: Simplify the right side of the equation:

3·4x−1= 27

Step 3: Divide both sides by 3 to isolate 4x−1:

4x−1= 9

Step 4: Rewrite the equation using the equivalent exponential form:

x−1 = log4(9)

Step 5: Use the definition of a logarithm to rewrite the equation as an

exponential form:

4x−1= 9

Step 6: Since 9 = 32, rewrite the equation using the same base:

4x−1= 42

Step 7: Set the exponents equal to each other:

x−1 = 2

Step 8: Add 1 to both sides to solve for x:

x= 3

Therefore, the solution to the equation is x= 3.

Question 12

Question

Solve the equation 2x= 5 −3xfor x.

Solution

Step 1: Rewrite the equation as 2x+ 3x= 5.

Step 2: Since 2xand 3xare both positive for all x, it follows that 2x+ 3x>0

for all x.

Step 3: Observe that x= 2 is a solution to the equation since 22+ 32=

4 + 9 = 13 = 5.

Step 4: To prove that x= 2 is the only solution, we will calculate the

derivatives of both sides of the equation.

Step 5: Let f(x) = 2x+ 3x, and find f′(x).

Step 6: f′(x) = ln(2) ·2x+ ln(3) ·3x.

Step 7: Let g(x) = 5, and find g′(x).

Step 8: g′(x) = 0.

Step 9: Comparing f′(x)and g′(x), it is clear that f′(x)=g′(x)for all x,

which means x= 2 is the only solution to the equation 2x+ 3x= 5.

Therefore, the solution to the equation 2x= 5 −3xis x= 2.

Question 13

Question

Solve the exponential equation 3x−2= 81 for x.

Solution

Step 1: Rewrite 81 as a power of 3.

81 = 34

Step 2: Substitute 34for 81 in the original equation and simplify.

3x−2= 34

x−2 = 4

Step 3: Solve for xby adding 2 to both sides of the equation.

x= 4 + 2

x= 6

Step 4: Check the solution by substituting x= 6 into the original equation.

36−2= 81

34= 81

81 = 81

Therefore, the solution to the equation 3x−2= 81 is x= 6.

Question 14

Question

Solve for xin the equation 23x= 5.

Solution

Step 1: Take the natural logarithm of both sides to eliminate the exponential

function.

ln(23x)= ln(5)

Step 2: Apply the logarithmic property ln(ab)=bln(a)to simplify the left

side.

3xln(2) = ln(5)

Step 3: Divide both sides by 3ln(2) to solve for x.

x=ln(5)

3 ln(2)

Step 4: Use the property logb(a) = ln(a)

ln(b)to rewrite xin a more simplified

form.

x=ln(5)

3 ln(2) =loge(5)

3 loge(2)

So, the solution to the equation 23x= 5 is x=loge(5)

3 loge(2) .

Question 15

Question

Samantha invested $5000 in a high yield savings account with an annual interest

rate of 4.5

Solution

Step 1: First, we identify the formula for continuous compound interest:

A=P·ert

where: - Ais the amount of money accumulated after tyears, - Pis the principal

amount (initial investment), - ris the annual interest rate (in decimal form),

and - tis the time the money is invested for.

Step 2: Plug in the given values into the formula:

A= 5000 ·e0.045×10

Step 3: Calculate the amount after 10 years:

A= 5000 ·e0.45

Step 4: Use the approximation e≈2.71828 to evaluate the expression:

A≈5000 ·2.718280.45

Step 5: Compute the final result:

A≈5000 ·1.56934 ≈7846.7

Therefore, Samantha will have approximately $7846.7 in her account after

10 years.

Question 16

Question

Solve the exponential equation for x:52x+1 = 125.

Solution

Step 1: Rewrite 125 as a power of 5.

125 = 53

Step 2: Set the exponents equal to each other and solve for x.

2x+ 1 = 3

2x= 2

x= 1

Step 3: Check the solution by substituting x= 1 back into the original

equation.

52(1)+1 = 53

53= 125

Therefore, the solution to the exponential equation is x= 1.

Question 17

Question

Let f(x) = log2(x2−1) be a logarithmic function. Find the domain of the

function f(x).

Solution

Step 1: The domain of a logarithmic function f(x) = logb(u(x)) is the set of

all real numbers xsuch that u(x)>0. Here, u(x) = x2−1and the base of the

logarithm is b= 2.

Step 2: Set the expression inside the logarithm greater than zero and solve

for x:

x2−1>0

Step 3: Factor the quadratic inequality:

(x−1)(x+ 1) >0

Step 4: The critical points are x=−1and x= 1. We create a sign chart

to analyze where the inequality is true:

x < −1−1< x < 1x > 1

x−1− − +

x+ 1 −+ +

(x−1)(x+ 1) + −+

Step 5: The inequality (x−1)(x+ 1) >0is true when x < −1or x > 1.

Therefore, the domain of f(x)is (−∞,−1) ∪(1,∞).

Question 18

Question

Solve the exponential equation 32x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3:

27 = 33

Step 2: Substitute 33for 27 in the equation:

32x−1= 33

Step 3: Since the bases are the same, we can set the exponents equal to each

other:

2x−1 = 3

Step 4: Solve for x:

2x−1 = 3

2x= 4

x= 2

Step 5: Check the solution by substituting x= 2 back into the original

equation:

32(2)−1= 33

33= 33

Since the equation holds true, the solution x= 2 is correct.

Question 19

Question

Solve the exponential equation 32x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3.

27 = 33

Step 2: Set the exponential equation in terms of the base 3.

32x−1= 33

Step 3: Equate the exponents.

2x−1 = 3

Step 4: Solve for x.

2x= 4

x= 2

Step 5: Check the solution. Substitute x= 2 back into the original equation:

32(2)−1= 33

33= 27

Since the left side equals the right side, x= 2 is the correct solution to the

exponential equation 32x−1= 27.

Question 20

Question

A certain radioactive substance decays over time. The amount of the substance

present after tyears is given by the function A(t) = A0e−kt, where A0is the

initial amount of the substance and kis a positive constant. If 30

Solution

Step 1: Given that 30

A(4) = A0e−4k= 0.7A0

Step 2: Divide both sides of the equation by A0to simplify the equation.

e−4k= 0.7

Step 3: Take the natural logarithm of both sides to solve for k.

ln(e−4k)= ln(0.7)

−4k= ln(0.7)

Step 4: Solve for kby dividing both sides by −4.

k=ln(0.7)

−4=−ln(7)

Therefore, the value of kis −ln(7)

Question 21

Question

Solve the exponential equation 23x+1 −2x+2 = 8.

Solution

Step 1: We can rewrite 8as 23. Therefore, the equation becomes 23x+1 −2x+2 =

23.

Step 2: By using the properties of exponents, we can rewrite 23x+1 as 2·23x

and 2x+2 as 2·2x. Substituting these into the equation gives us 2·23x−2·2x= 23.

Step 3: Factor out a 2xterm from both terms on the left side, giving us

2x(2 ·22x−2) = 23.

Step 4: Simplify the expression inside the parentheses to get 2x(4 ·2x−2) =

23.

Step 5: Further simplifying, we have 2x(2x+2 −2) = 23.

Step 6: Applying the property of exponents again, we find that 2x+2 = 4·2x.

Substituting this back into the equation gives us 2x(4 ·2x−2) = 23.

Step 7: Solve for 2xby dividing both sides by (4 ·2x−2):2x=23

4·2x−2.

Step 8: Substitute 2x=yto simplify the equation to y=8

4y−2.

Step 9: Multiply both sides by (4y−2) to get y(4y−2) = 8.

Step 10: Expand and rearrange the equation to get 4y2−2y−8 = 0.

Step 11: Solve this quadratic equation by factoring or using the quadratic

formula to find the value(s) of y.

Step 12: Once you have found the value(s) of y, substitute back to find the

corresponding values of 2x. Remember to check for extraneous solutions.

Question 22

Question

A certain population of bacteria doubles every 3 hours. If there are 100 initial

bacteria, how many bacteria will there be after 12 hours?

Solution

Step 1: Determine the rate of growth for the bacteria population.

Given that the population doubles every 3 hours, we can express this ex-

ponential growth using the formula for exponential growth: A(t) = A0·(2) t

where: - A(t)is the population after thours, - A0is the initial population, - 2

is the factor by which the population doubles every 3 hours, and - t

3represents

the number of 3-hour intervals.

Step 2: Substitute the given values into the formula.

Given that the initial population is 100 and we want to find the population

after 12 hours, we have: A(12) = 100 ·(2)12

Step 3: Solve for the final population after 12 hours.

A(12) = 100 ·24

A(12) = 100 ·16

A(12) = 1600

After 12 hours, there will be 1600 bacteria in the population.

Question 23

Question

A population of bacteria triples every 5 minutes in a lab experiment. If there

were initially 100 bacteria, how long will it take for the population to reach 5000

bacteria?

Solution

Step 1: Let’s denote the initial number of bacteria as P0and the time it takes

for the population to reach a certain number of bacteria as t. Given that the

population triples every 5 minutes, we can express the population growth as an

exponential function: P(t) = P0·3t/5.

Step 2: We are given that P0= 100 bacteria. Substituting this value into

the exponential function, we have P(t) = 100 ·3t/5.

Step 3: We also know that we want to find the time twhen the population

reaches 5000 bacteria. So, we have the equation 5000 = 100 ·3t/5.

Step 4: Dividing both sides by 100, we get 50 = 3t/5.

Step 5: Taking the natural logarithm of both sides, we have ln(50) =

ln(3t/5).

Step 6: Using the properties of logarithms, we can rewrite the right side as

t/5·ln(3).

Step 7: Simplifying, we find t= 5 ·ln(50)

ln(3) .

Step 8: Using a calculator to approximate, we find t≈24.57 minutes. There-

fore, it will take approximately 24.57 minutes for the population to reach 5000

bacteria.

Question 24

Question

Solve the exponential equation: 32x−1= 9.

Solution

Step 1: Rewrite both sides of the equation with the same base.

32x−1= 9

32x−1= 32

Step 2: Since the bases are equal, we can set the exponents equal to one

another.

2x−1 = 2

Step 3: Solve for x.

2x−1 = 2

2x= 3

x=3

x= 1.5

Step 4: Therefore, the solution to the exponential equation 32x−1= 9 is

x= 1.5.

Question 25

Question

Solve the exponential equation 32x−1= 27.

Solution

Step 1: Rewrite 27 as a power of 3:

27 = 33

Step 2: Substitute 33for 27 in the equation:

32x−1= 33

Step 3: Since the bases are the same, set the exponents equal to each other:

2x−1 = 3

Step 4: Solve for xby isolating x:

2x= 4

x= 2

Step 5: Check the solution by substituting x= 2 back into the original

equation:

32(2)−1= 33

33= 33

Step 6: Since the left side equals the right side, x= 2 is the correct solution.

Question 26

Question

Suppose a radioactive substance decays according to the formula A(t) = A0e−kt,

where A(t)is the amount of substance remaining after tdays, A0is the initial

amount of substance, and kis a constant. If 100 grams of a substance decay to

70 grams in 10 days, find the value of k.

Solution

Step 1: We are given that A0= 100 grams, A(10) = 70 grams, and t= 10 days.

Substituting these values into the formula A(t) = A0e−kt, we get:

70 = 100e−10k

Step 2: Divide both sides by 100 to isolate the exponential term:

100 =e−10k

Step 3: Simplify the left side:

0.7 = e−10k

Step 4: Take the natural logarithm of both sides to solve for k:

ln(0.7) = ln(e−10k)

ln(0.7) = −10kln(e)

ln(0.7) = −10k

Step 5: Solve for kby dividing both sides by −10:

k=ln(0.7)

−10

Therefore, the value of kis k=ln(0.7)

−10 .