Exponential Functions and Equations Practice Questions — SAT — Page 2

If $4^x = 5$ , what is the value of $2^{4x - 1}$ ?

Show answer & explanation

Answer: 12.5

Answer

12.5

By writing

4^x

as

(2^2)^x = 2^{2x}

, we find that

2^{2x} = 5

. The expression

2^{4x-1}

can be rewritten using exponent properties as

\frac{2^{4x}}{2^1} = \frac{(2^{2x})^2}{2}

. Substituting

2^{2x} = 5

into this expression yields

\frac{5^2}{2} = \frac{25}{2} = 12.5

. Thus, the correct numerical response is 12.5.

Step-by-Step Solution

1

Express the given equation in terms of base 2.

2^{2x} = 5

Since

4 = 2^2

, we can rewrite

4^x

as

(2^2)^x

. Applying the power of a power rule,

(a^m)^n = a^{mn}

, gives

(2^2)^x = 2^{2x}

.

2

Rewrite the expression to be evaluated using exponent properties.

2^{4x - 1} = \frac{(2^{2x})^2}{2}

Using the division property of exponents,

a^{m-n} = \frac{a^m}{a^n}

, we can write

2^{4x - 1}

as

\frac{2^{4x}}{2}

. Then, using the power of a power rule in reverse,

2^{4x} = (2^{2x})^2

.

3

Substitute the known value of

2^{2x}

and simplify the numerical expression.

12.5

Substitute

2^{2x} = 5

into the expression

\frac{(2^{2x})^2}{2}

to obtain

\frac{5^2}{2} = \frac{25}{2}

, which simplifies to

12.5

.

Key Concept

Manipulating exponential equations by expressing bases in terms of common prime factors and applying properties of exponents.

Estimated Time:1m 30s

Question 22Question →

An exponential function $g$ is defined by $g(x) = p \cdot q^x$ , where $p$ and $q$ are positive constants. If $g\left(\frac{1}{2}\right) = 6$ and $g\left(\frac{3}{2}\right) = 54$ , what is the value of $g(2)$ ?

Show answer & explanation

Answer: 162

Answer

162

The correct answer is 162. Substituting the coordinates into the function gives the system

p \cdot q^{1/2} = 6

and

p \cdot q^{3/2} = 54

. Dividing the second equation by the first yields

q = 9

. Substituting

q = 9

into the first equation yields

p \cdot 3 = 6

, so

p = 2

. The complete function is

g(x) = 2 \cdot 9^x

, and evaluating

g(2)

gives

2 \cdot 81 = 162

.

Step-by-Step Solution

1

Substitute the given values into the function definition to set up a system of equations.

p \cdot q^{1/2} = 6

and

p \cdot q^{3/2} = 54

To represent the given coordinate relationships using the exponential model parameters.

2

Divide the equation for

g\left(\frac{3}{2}\right)

by the equation for

g\left(\frac{1}{2}\right)

to eliminate

p

and solve for

q

.

q = 9

Using the exponent quotient rule,

\frac{q^{3/2}}{q^{1/2}} = q^{3/2 - 1/2} = q^1 = q

. Dividing the constants gives

\frac{54}{6} = 9

.

3

Substitute

q = 9

back into the first equation and solve for

p

.

p = 2

Evaluating the fractional exponent gives

9^{1/2} = \sqrt{9} = 3

. Substituting this in yields

3p = 6

, which simplifies to

p = 2

.

4

Evaluate the completed exponential function

g(x) = 2 \cdot 9^x

at

x = 2

.

162

Calculating

g(2) = 2 \cdot 9^2 = 2 \cdot 81 = 162

solves for the target value.

Key Concept

Solving a system of exponential equations using fractional exponent rules and base evaluation.

Question 23Question →

A radioactive substance decays according to the equation $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{8}}$ , where $N(t)$ is the amount of the substance remaining after $t$ days, and $N_0$ is the initial amount. If the substance decays to $\frac{1}{16}$ of its initial amount in $d$ days, what is the value of $d$ ?

32

4

2

128

Show answer & explanation

Answer: 32

Answer

32

The correct answer is 32. The remaining amount of the substance is given as

\frac{1}{16}

of the initial amount, so

\frac{N(d)}{N_0} = \frac{1}{16}

. Substituting

d

for

t

in the decay formula yields

\left(\frac{1}{2}\right)^{\frac{d}{8}} = \frac{1}{16}

. Since

\frac{1}{16} = \left(\frac{1}{2}\right)^4

, we can set the exponents equal to each other:

\frac{d}{8} = 4

. Solving for

d

gives

d = 32

.

Step-by-Step Solution

1

Set up the equation for the remaining fraction of the substance.

\frac{N(d)}{N_0} = \frac{1}{16}

The problem states the substance decays to

\frac{1}{16}

of its initial amount

N_0

in

d

days.

2

Substitute the given decay formula into the equation.

\left(\frac{1}{2}\right)^{\frac{d}{8}} = \frac{1}{16}

By replacing

t

with

d

in the formula

\frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{8}}

, we get the expression for the remaining fraction.

3

Rewrite

\frac{1}{16}

as a power with base

\frac{1}{2}

to solve for the exponent.

\left(\frac{1}{2}\right)^4 = \frac{1}{16}

Since

2^4 = 16

, the fraction

\frac{1}{16}

is equivalent to

\left(\frac{1}{2}\right)^4

.

4

Equate the exponents and solve for

d

.

\frac{d}{8} = 4 \implies d = 32

Since the bases are equal, their exponents must be equal. Multiplying both sides by 8 isolates

d

.

Key Concept

Solving exponential equations by equating bases

Question 24Question →

In the system of equations below, $x$ and $y$ are real numbers.

4^x \cdot 8^y = 128

9^x \cdot 3^y = 243

What is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

To solve the system of equations, express all terms in each equation using common bases. In the first equation,

4^x \cdot 8^y = 128

can be written as

(2^2)^x \cdot (2^3)^y = 2^7

, which simplifies to

2^{2x+3y} = 2^7

, meaning

2x + 3y = 7

. In the second equation,

9^x \cdot 3^y = 243

can be written as

(3^2)^x \cdot 3^y = 3^5

, which simplifies to

3^{2x+y} = 3^5

, meaning

2x + y = 5

. Subtracting the two equations gives

(2x + 3y) - (2x + y) = 7 - 5 \implies 2y = 2 \implies y = 1

. Substituting

y = 1

back into

2x + y = 5

gives

2x + 1 = 5 \implies 2x = 4 \implies x = 2

.

Step-by-Step Solution

1

Rewrite the first equation using a common base of 2.

2x + 3y = 7

By converting

4^x

to

(2^2)^x = 2^{2x}

,

8^y

to

(2^3)^y = 2^{3y}

, and

128

to

2^7

, we can equate the exponents:

2x + 3y = 7

.

2

Rewrite the second equation using a common base of 3.

2x + y = 5

By converting

9^x

to

(3^2)^x = 3^{2x}

, keeping

3^y

, and converting

243

to

3^5

, we can equate the exponents:

2x + y = 5

.

3

Solve the system of linear equations for

y

.

y = 1

Subtracting

2x + y = 5

from

2x + 3y = 7

eliminates

x

, leaving

2y = 2

, which gives

y = 1

.

4

Substitute

y = 1

into one of the linear equations to solve for

x

.

x = 2

Substituting

y = 1

into

2x + y = 5

yields

2x + 1 = 5

, which simplifies to

2x = 4

, so

x = 2

.

Key Concept

Solving systems of exponential equations by converting to a common base and applying exponent laws.

Question 25Question →

If $27^{2x - 1} \cdot 9^{y + 2} = 3^{13}$ and $8^{x - y} = 2^{y - 9}$ , what is the value of $x + y$ ?

2

4

6

9

Show answer & explanation

Answer: 4

Answer

4

To solve the system, we convert all exponential terms to their prime bases. The first equation becomes

3^{6x - 3} \cdot 3^{2y + 4} = 3^{13}

, which simplifies to

3^{6x + 2y + 1} = 3^{13}

. Equating the exponents yields

6x + 2y + 1 = 13

, which simplifies to

3x + y = 6

. The second equation becomes

2^{3x - 3y} = 2^{y - 9}

. Equating these exponents yields

3x - 4y = -9

. Solving this system of equations gives

x = 1

and

y = 3

. The sum of these values is

1 + 3 = 4

.

Step-by-Step Solution

1

Convert the bases of the first equation to a common base of 3.

Since

27 = 3^3

and

9 = 3^2

, we can rewrite the equation

27^{2x - 1} \cdot 9^{y + 2} = 3^{13}

as

(3^3)^{2x - 1} \cdot (3^2)^{y + 2} = 3^{13}

. Applying the power of a power rule, this simplifies to

3^{6x - 3} \cdot 3^{2y + 4} = 3^{13}

. Using the product rule of exponents, we add the exponents to get

3^{6x + 2y + 1} = 3^{13}

.

This allows us to equate the exponents and form a linear equation.

2

Set the exponents equal to each other to obtain the first linear equation.

6x + 2y + 1 = 13 \implies 6x + 2y = 12 \implies 3x + y = 6

.

Since the bases are identical on both sides, their exponents must be equal.

3

Convert the bases of the second equation to a common base of 2 and equate their exponents.

Since

8 = 2^3

, rewrite

8^{x - y} = 2^{y - 9}

as

(2^3)^{x - y} = 2^{y - 9} \implies 2^{3x - 3y} = 2^{y - 9}

. Equating the exponents gives

3x - 3y = y - 9 \implies 3x - 4y = -9

.

This gives us a second linear equation to form a system of equations.

4

Solve the system of equations for the variables.

From the first equation, we have

3x = 6 - y

. Substituting this into the second equation gives

(6 - y) - 4y = -9 \implies 6 - 5y = -9 \implies -5y = -15 \implies y = 3

. Substituting

y = 3

back into the first equation yields

3x + 3 = 6 \implies 3x = 3 \implies x = 1

.

Solving the system of linear equations provides the individual values of the variables.

5

Calculate the sum of the variables.

x + y = 1 + 3 = 4

.

This answers the question asking for the sum of the variables.

Key Concept

Expressing exponential terms with a common base to form and solve a system of linear equations.

Question 26Question →

A certain substance decays radioactively such that the mass of the substance, in grams, remaining after $t$ days is modeled by the function $M(t) = 80 \cdot 2^{-\frac{t}{5}}$ . After how many days will the mass of the substance be $10$ grams?

Show answer & explanation

Answer: 15

Answer

15

To find the number of days after which the mass of the substance is

10

grams, we substitute

10

for

M(t)

in the given function, yielding

10 = 80 \cdot 2^{-\frac{t}{5}}

. Dividing both sides by

80

gives

\frac{10}{80} = 2^{-\frac{t}{5}}

, which simplifies to

\frac{1}{8} = 2^{-\frac{t}{5}}

. Since

\frac{1}{8}

can be written as

2^{-3}

, the equation becomes

2^{-3} = 2^{-\frac{t}{5}}

. Because the bases are the same, we equate the exponents:

-3 = -\frac{t}{5}

. Multiplying both sides by

-5

gives

t = 15

.

Step-by-Step Solution

1

Set the mass

M(t)

equal to

10

in the given function.

10 = 80 \cdot 2^{-\frac{t}{5}}

We want to find the value of

t

when the remaining mass of the substance is

10

grams.

2

Divide both sides of the equation by

80

.

\frac{1}{8} = 2^{-\frac{t}{5}}

To isolate the exponential expression.

3

Express both sides of the equation with a common base of

2

.

2^{-3} = 2^{-\frac{t}{5}}

Since

8 = 2^3

, the fraction

\frac{1}{8}

can be written as

2^{-3}

. Having the same base on both sides allows us to equate the exponents.

4

Set the exponents equal to each other and solve for

t

.

t = 15

Because the bases are equal, the exponents must be equal:

-3 = -\frac{t}{5}

.

Key Concept

Solving exponential equations by finding a common base.

Question 27Question →

If $3^{2x - 1} \cdot 9^{y + 1} = 27^{x + y}$ , which of the following equations expresses $x$ in terms of $y$ ?

x = 1 - y

x = y + 1

x = -y

x = -y - 1

Show answer & explanation

Answer: $x = 1 - y$

Answer

The equation expressing

x

in terms of

y

is

x = 1 - y

.

The correct option is

x = 1 - y

. To solve the equation, each term is rewritten with base 3:

3^{2x-1} \cdot (3^2)^{y+1} = (3^3)^{x+y}

. Using the exponent rule

(b^m)^n = b^{mn}

, this becomes

3^{2x-1} \cdot 3^{2y+2} = 3^{3x+3y}

. Using the rule

b^m \cdot b^n = b^{m+n}

, the left side simplifies to

3^{2x+2y+1}

. Equating the exponents gives

2x+2y+1 = 3x+3y

. Subtracting

2x

and

2y

from both sides results in

1 = x+y

, which isolated for

x

gives

x = 1-y

.

Step-by-Step Solution

1

Rewrite all bases in the equation as powers of 3.

Since

9 = 3^2

and

27 = 3^3

, the equation becomes

3^{2x - 1} \cdot (3^2)^{y + 1} = (3^3)^{x + y}

.

To apply exponent rules and combine terms, all bases must be identical.

2

Apply the power of a power rule

(b^m)^n = b^{mn}

to simplify the exponents.

3^{2x - 1} \cdot 3^{2y + 2} = 3^{3x + 3y}

Multiplying the outer exponent by the inner exponent simplifies the terms on both sides.

3

Apply the product of powers rule

b^m \cdot b^n = b^{m+n}

to combine the terms on the left side of the equation.

3^{(2x - 1) + (2y + 2)} = 3^{3x + 3y}

, which simplifies to

3^{2x + 2y + 1} = 3^{3x + 3y}

.

Adding exponents when multiplying expressions with the same base allows the left side to be represented as a single exponential term.

4

Set the exponents equal to each other and solve for

x

in terms of

y

.

2x + 2y + 1 = 3x + 3y \implies 1 = x + y \implies x = 1 - y

If two exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Key Concept

Solving exponential equations by expressing terms with a common base and applying exponent laws.

Estimated Time:1m 30s

Question 28Question →

A scientist models the population of a bacteria culture using the function $P(t) = P_0 \cdot b^t$ , where $P(t)$ is the population $t$ hours after the start of the experiment, $P_0$ is the initial population, and $b$ is a constant. The table below shows the population at two different times:

$t$ (hours)	$P(t)$
$2$	$1,800$
$5$	$48,600$

If the population of the bacteria culture is $1,312,200$ after $k$ hours, what is the value of $k$ ?

4

8

9

14

Show answer & explanation

Answer: 8

Answer

8

The correct value is 8. By setting up the ratio of the population at

t = 5

to

t = 2

, we find that

\frac{P(5)}{P(2)} = b^3 = \frac{48,600}{1,800} = 27

, which yields an hourly growth factor of

b = 3

. Using

P(2) = P_0 \cdot 3^2 = 1,800

, we determine the initial population

P_0

is 200. To find the hour

k

when the population is

1,312,200

, we solve

200 \cdot 3^k = 1,312,200

, which simplifies to

3^k = 6,561

. Since

3^8 = 6,561

, we find

k = 8

. Alternatively, we can calculate

\frac{P(k)}{P(5)} = \frac{1,312,200}{48,600} = 27 = 3^3

, meaning the population triples 3 more times after

t = 5

, giving

k = 5 + 3 = 8

.

Step-by-Step Solution

1

Set up the ratio of the population at

t = 5

to the population at

t = 2

to find the growth factor

b

.

\frac{P(5)}{P(2)} = \frac{P_0 \cdot b^5}{P_0 \cdot b^2} = b^3 = \frac{48,600}{1,800} = 27

Dividing the function values eliminates the initial population

P_0

and isolates the base

b

.

2

Solve for the growth factor

b

.

b = 3

Since

b^3 = 27

, taking the cube root of both sides gives

b = 3

.

3

Find the initial population

P_0

using

P(2) = 1,800

and

b = 3

.

1,800 = P_0 \cdot 3^2 \Rightarrow 1,800 = 9P_0 \Rightarrow P_0 = 200

Substituting the known values into the function formula allows us to solve for

P_0

.

4

Set up the equation for the population after

k

hours and solve for

k

.

1,312,200 = 200 \cdot 3^k \Rightarrow 3^k = \frac{1,312,200}{200} = 6,561

Dividing both sides by 200 isolates the exponential term

3^k

.

5

Find the exponent

k

by expressing 6,561 as a power of 3.

3^k = 3^8 \Rightarrow k = 8

Since

3^8 = 6,561

, the exponents must be equal, so

k = 8

.

Key Concept

Determining parameters of an exponential growth function from given data points and using the function to solve for an unknown time variable.

Alternative Method

Instead of solving for the initial population

P_0

, we can compare the target population of

1,312,200

to the population at

t = 5

. Since the population triples every hour (

b = 3

), the ratio of the population at

t = k

to the population at

t = 5

is

3^{k-5}

. We have

\frac{1,312,200}{48,600} = 27

, which is

3^3

. Therefore,

3^{k-5} = 3^3

, which directly gives

k - 5 = 3

, or

k = 8

.

Estimated Time:1m 15s

Question 29Question →

The value of a certain car model decreases by $12\%$ each year. If the initial value of the car is $\$25,000$ , which of the following functions models the value of the car, $V(t)$ , in dollars, $t$ years after it was purchased?

V(t) = 25,000 - 0.12t

V(t) = 25,000(0.12)^t

V(t) = 25,000(0.88)^t

V(t) = 25,000(1.12)^t

Show answer & explanation

Answer: $V(t) = 25,000(0.88)^t$

Answer

The function

V(t) = 25,000(0.88)^t

The initial value of the car is

25,000

. An annual decrease of

12\%

means that the car retains

100\% - 12\% = 88\%

of its value from the previous year. This results in a decay factor of

1 - 0.12 = 0.88

. Substituting these values into the exponential model

V(t) = a(b)^t

yields

V(t) = 25,000(0.88)^t

.

Step-by-Step Solution

1

Identify the initial value of the car.

The initial value is

25,000

.

This is the value of the car at

t = 0

, which corresponds to the coefficient

a

in the general exponential model

V(t) = a(b)^t

.

2

Determine the growth or decay factor base,

b

.

The base is

1 - 0.12 = 0.88

.

Since the car's value decreases by

12\%

each year, it retains

100\% - 12\% = 88\%

of its value. This is represented as a decay factor of

0.88

.

3

Formulate the exponential decay equation.

The equation is

V(t) = 25,000(0.88)^t

.

By substituting the initial value

a = 25,000

and the decay base

b = 0.88

into

V(t) = a(b)^t

.

Key Concept

Exponential decay models and interpretation of growth/decay factors

Alternative Method

Calculate the value of the car after one year (

t = 1

). A

12\%

decrease on a

\$25,000

car means its value drops by

\$3,000

to

\$22,000

. Substituting

t=1

into the correct function should yield

22,000

. Calculating for

V(t) = 25,000(0.88)^t

at

t = 1

gives

25,000(0.88)^1 = 22,000

, confirming the choice.

Estimated Time:45s

Question 30Question →

The population of a colony of bacteria doubles every $15$ minutes. The population of the colony, $P(t)$ , after $t$ hours can be modeled by the function $P(t) = P_0 \cdot 8^{kt}$ , where $P_0$ is the initial population of the colony and $k$ is a constant. What is the value of $k$ ?

\frac{1}{12}

\frac{3}{4}

\frac{4}{3}

12

Show answer & explanation

Answer: $\frac{4}{3}$

Answer

The correct value of

k

is

\frac{4}{3}

.

Since the population doubles every

15

minutes, it doubles

4

times per hour. Therefore, the population after

t

hours is

P_0 \cdot 2^{4t}

. To express this model in the form

P_0 \cdot 8^{kt}

, we equate the growth factors:

2^{4t} = 8^{kt}

. Since

8 = 2^3

, this can be rewritten as

2^{4t} = (2^3)^{kt} = 2^{3kt}

. Setting the exponents equal gives

4t = 3kt

, which simplifies to

4 = 3k

and yields

k = \frac{4}{3}

.

Step-by-Step Solution

1

Determine the number of doubling periods per hour.

The population doubles

4

times per hour.

Since there are

60

minutes in an hour and the population doubles every

15

minutes, the number of doubling periods in one hour is

\frac{60}{15} = 4

.

2

Formulate the population growth equation in terms of base 2.

P(t) = P_0 \cdot 2^{4t}

If the population doubles

4

times per hour, then after

t

hours, it has doubled

4t

times, representing a growth factor of

2^{4t}

.

3

Equate the growth model to the given function and express using a common base.

2^{4t} = (2^3)^{kt} \implies 2^{4t} = 2^{3kt}

Set the two expressions for the population equal:

P_0 \cdot 2^{4t} = P_0 \cdot 8^{kt}

. Dividing by

P_0

gives

2^{4t} = 8^{kt}

. Since

8 = 2^3

, rewrite the right side to have a base of

2

.

4

Solve for the constant

k

by equating the exponents.

4t = 3kt \implies 4 = 3k \implies k = \frac{4}{3}

Since the bases on both sides of the equation are equal, their exponents must also be equal. Divide both sides by

t

(since

t > 0

) and isolate

k

.

Key Concept

Converting bases in exponential functions and adjusting exponents for different units of time.

Question 31Question →

If $5^{a + 2} = 25^{b}$ and $a + b = 10$ , what is the value of $b$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

The correct answer is 4. By rewriting

25^b

as

(5^2)^b = 5^{2b}

, we can set the exponents equal to get

a + 2 = 2b

. Substituting

a = 10 - b

from the second equation into this relation yields

(10 - b) + 2 = 2b

. Simplifying this gives

12 = 3b

, which results in

b = 4

.

Step-by-Step Solution

1

Rewrite the exponential equation using a common base.

5^{a + 2} = 5^{2b}

Since

25 = 5^2

, we write

25^b

as

(5^2)^b = 5^{2b}

to match the base on the left side.

2

Equate the exponents from the base 5 terms.

a + 2 = 2b

When bases are equal, their exponents must be equal for the equation to hold.

3

Express

a

in terms of

b

using the linear equation

a + b = 10

.

a = 10 - b

This allows for substitution into the equation from Step 2 to solve for a single variable.

4

Substitute

a = 10 - b

into

a + 2 = 2b

and solve for

b

.

b = 4

Substituting gives

(10 - b) + 2 = 2b \implies 12 - b = 2b \implies 12 = 3b \implies b = 4

.

Key Concept

Solving systems of equations involving exponential equations by finding a common base.

Question 32Question →

The population of a certain species of wildflower in a national park increases by $300\%$ every $8$ years. The population can be modeled by the function $P(t) = P_0(1 + r)^t$ , where $P_0$ is the initial population, $r$ is the constant annual growth rate, and $t$ is the time in years. Which of the following is closest to the value of $r$ ?

0.15

0.19

0.38

1.19

Show answer & explanation

Answer: 0.19

Answer

The correct value of the annual growth rate is

0.19

.

An increase of

300\%

means the population becomes

100\% + 300\% = 400\%

of its initial value, which corresponds to a growth factor of

4

over an

8

-year period. The annual growth rate

r

satisfies the equation

(1 + r)^8 = 4

. Solving for

1 + r

gives

1 + r = 4^{1/8} = (2^2)^{1/8} = 2^{1/4} \approx 1.1892

. Subtracting

1

from both sides yields the annual growth rate

r \approx 0.1892

, which is closest to

0.19

.

Step-by-Step Solution

1

Determine the total growth factor over the 8-year period from the percentage increase.

The population increases by

300\%

, meaning the new population is

100\% + 300\% = 400\%

of the initial population. This corresponds to a growth factor of

4

every

8

years.

An increase of

300\%

adds

3

times the original value to the original value, resulting in a total of

4

times the original value.

2

Set up the equation relating the annual growth rate

r

to the 8-year growth factor.

(1+r)^8 = 4

Since the population is multiplied by

1+r

each year, after

8

years it is multiplied by

(1+r)^8

, which must equal the total growth factor of

4

.

3

Solve the equation for

r

and round to the nearest hundredth.

1+r = 4^{1/8} \approx 1.1892 \implies r \approx 0.1892 \approx 0.19

Taking the eighth root of both sides gives the annual growth factor

1+r \approx 1.19

. Subtracting

1

isolates the annual growth rate

r

.

Key Concept

Relating exponential growth rates and growth factors across different time periods

Alternative Method

Instead of solving

(1+r)^8 = 4

, we can write the function directly using the 8-year growth factor as

P(t) = P_0 \cdot 4^{t/8}

. Rewriting this in the form

P(t) = P_0(4^{1/8})^t

shows that the annual growth factor is

4^{1/8}

. We can approximate

4^{1/8} \approx 1.1892

, which corresponds to an annual growth rate

r = 1.1892 - 1 = 0.1892

, or approximately

0.19

.

Estimated Time:1m 30s

Question 33Question →

If $4^{2x + 1} = 8^{x - 1}$ , what is the value of $x$ ?

-5

-2

-1

5

Show answer & explanation

Answer: -5

Answer

-5

The correct answer is

-5

. By converting the bases of both sides to

2

, the equation becomes

2^{2(2x+1)} = 2^{3(x-1)}

. Equating the exponents gives

2(2x+1) = 3(x-1)

, which expands to

4x+2 = 3x-3

. Solving for

x

yields

x = -5

.

Step-by-Step Solution

1

Express both bases as powers of 2

4 = 2^2

and

8 = 2^3

To solve an exponential equation, we need to rewrite the terms with a common base.

2

Substitute the bases back into the equation and apply exponent power rules

(2^2)^{2x + 1} = (2^3)^{x - 1} \Rightarrow 2^{2(2x + 1)} = 2^{3(x - 1)}

Applying the power of a power property,

(b^m)^n = b^{mn}

, allows us to simplify the exponents.

3

Set the exponents equal to each other

2(2x + 1) = 3(x - 1)

Since the bases are identical and positive, their exponents must be equal.

4

Distribute and solve the linear equation for

x

4x + 2 = 3x - 3 \Rightarrow x = -5

Distributing the constants gives

4x + 2 = 3x - 3

. Subtracting

3x

from both sides gives

x + 2 = -3

, and subtracting

2

from both sides isolates the variable to find

x = -5

.

Key Concept

Solving exponential equations by expressing both sides with a common base

Question 34Question →

If $9^{y - 1} \cdot 3^{2y} = 27^{y + 2}$ , what is the value of $y$ ?

1.5

4

7

8

Show answer & explanation

Answer: 8

Answer

8

To solve the equation, we rewrite each term using the common base of

3

. Since

9 = 3^2

and

27 = 3^3

, the equation

9^{y - 1} \cdot 3^{2y} = 27^{y + 2}

can be written as

(3^2)^{y-1} \cdot 3^{2y} = (3^3)^{y+2}

. Applying the power of a power rule, we get

3^{2(y-1)} \cdot 3^{2y} = 3^{3(y+2)}

, which simplifies to

3^{2y-2} \cdot 3^{2y} = 3^{3y+6}

. Using the product rule of exponents to combine the left side gives

3^{2y-2+2y} = 3^{4y-2} = 3^{3y+6}

. Setting the exponents equal to each other gives the linear equation

4y - 2 = 3y + 6

. Solving for

y

yields

y = 8

.

Step-by-Step Solution

1

Express all bases in the equation as powers of 3.

9^{y-1} = (3^2)^{y-1} = 3^{2y-2}

and

27^{y+2} = (3^3)^{y+2} = 3^{3y+6}

. The original equation becomes

3^{2y-2} \cdot 3^{2y} = 3^{3y+6}

.

Having a common base allows us to combine exponents using the properties of exponents.

2

Simplify the left side of the equation by adding the exponents.

3^{(2y-2) + 2y} = 3^{4y-2}

, so the equation is

3^{4y-2} = 3^{3y+6}

.

When multiplying exponential expressions with the same base, we add their exponents.

3

Set the exponents equal to each other and solve for

y

.

4y - 2 = 3y + 6 \implies y - 2 = 6 \implies y = 8

.

Since the bases are equal and positive (and not equal to 1), their exponents must be equal.

Key Concept

Solving exponential equations by expressing terms with a common base and applying exponent laws.

Estimated Time:1m 30s

Question 35Question →

If $\left(\frac{1}{8}\right)^{2x - 3} = 16^{x - 2}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1.7

Answer

1.7 (or 17/10)

The correct answer is 1.7 (or 17/10). By writing

\frac{1}{8}

as

2^{-3}

and

16

as

2^4

, the equation becomes

(2^{-3})^{2x - 3} = (2^4)^{x - 2}

. Applying exponent rules yields

2^{-6x + 9} = 2^{4x - 8}

. Equating the exponents gives

-6x + 9 = 4x - 8

, which simplifies to

10x = 17

, resulting in

x = 1.7

.

Step-by-Step Solution

1

Express both bases as powers of

2

.

(2^{-3})^{2x - 3} = (2^4)^{x - 2}

To solve the exponential equation, both sides must have a common base. Both

\frac{1}{8}

and

16

can be rewritten as powers of

2

.

2

Apply the power-of-a-power exponent rule

(a^m)^n = a^{mn}

to simplify the exponents.

2^{-3(2x - 3)} = 2^{4(x - 2)}

, which simplifies to

2^{-6x + 9} = 2^{4x - 8}

This allows us to write each side as a single exponential term with a base of

2

.

3

Set the exponents equal to each other.

-6x + 9 = 4x - 8

Since the bases are equal, their exponents must be equal for the equation to hold.

4

Solve the linear equation for

x

.

10x = 17

, which gives

x = 1.7

Isolate the variable

x

by algebraic manipulation.

Key Concept

Solving exponential equations by expressing terms with a common base and applying exponent rules.

Alternative Method

Alternatively, you can apply the definition of logarithms to both sides of the equation. Taking the logarithm base 2 of both sides gives

\log_2\left(\left(\frac{1}{8}\right)^{2x-3}\right) = \log_2\left(16^{x-2}\right)

. Using the power property of logarithms, this simplifies to

(2x - 3)\log_2\left(2^{-3}\right) = (x - 2)\log_2\left(2^4\right)

, which simplifies directly to

-3(2x - 3) = 4(x - 2)

. Solving this linear equation yields the same result,

x = 1.7

.

Estimated Time:1m 30s

Question 36Question →

For what value of $x$ is the equation $8^{x + 1} = 16^x$ true?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The correct answer is 3. To find the value of

x

, express both 8 and 16 as powers of 2:

8 = 2^3

and

16 = 2^4

. Substituting these values into the equation gives

(2^3)^{x + 1} = (2^4)^x

. Applying the power of a power rule,

(a^m)^n = a^{mn}

, yields

2^{3(x + 1)} = 2^{4x}

, which simplifies to

2^{3x + 3} = 2^{4x}

. Since the bases are now the same, their exponents must be equal:

3x + 3 = 4x

. Subtracting

3x

from both sides gives

x = 3

.

Step-by-Step Solution

1

Express the bases 8 and 16 as powers of 2.

(2^3)^{x + 1} = (2^4)^x

Writing both sides of the equation with a common base allows for the equating of exponents.

2

Apply the power of a power rule

(a^m)^n = a^{mn}

to simplify the exponents.

2^{3x + 3} = 2^{4x}

This simplifies each side of the equation to a single base with a single exponent.

3

Equate the exponents and solve the resulting linear equation.

3x + 3 = 4x

, which simplifies to

x = 3

Since the bases are equal, their exponents must be equal for the equation to hold true.

Key Concept

Solving exponential equations by expressing both sides with a common base and equating the exponents.

Alternative Method

An alternative approach is to write the equation in terms of base 4. While 8 is not an integer power of 4, we can write

8 = 4^{1.5}

and

16 = 4^2

. The equation becomes

(4^{1.5})^{x+1} = (4^2)^x

, which simplifies to

1.5(x+1) = 2x

. Solving this gives

1.5x + 1.5 = 2x \implies 0.5x = 1.5 \implies x = 3

.

Estimated Time:45s

Question 37Question →

For the exponential function $g(x) = k \cdot a^x$ , where $k$ and $a$ are positive constants, the value of $g(x + 1)$ is $20\%$ greater than the value of $g(x - 1)$ for all real numbers $x$ . If $g(2) = 15$ , what is the value of $g(6)$ ?

21.0

21.6

31.104

36.0

Show answer & explanation

Answer: 21.6

Answer

21.6

The correct value of 21.6 is obtained by setting up the ratio

g(x+1)/g(x-1) = a^2

. Since

g(x+1)

is

20\%

greater than

g(x-1)

, this ratio is equal to 1.20, so

a^2 = 1.20

. Since

g(6) = g(2) \cdot a^4 = 15 \cdot (a^2)^2

, substituting

a^2 = 1.20

gives

15 \cdot (1.20)^2 = 15 \cdot 1.44 = 21.6

.

Step-by-Step Solution

1

Express the values of the function at x + 1 and x - 1 using the definition of g(x).

g(x + 1) = k \cdot a^{x + 1}

and

g(x - 1) = k \cdot a^{x - 1}

To set up the mathematical relationship between the two values based on the function definition.

2

Apply the condition that g(x + 1) is 20% greater than g(x - 1).

k \cdot a^{x + 1} = 1.20 \cdot k \cdot a^{x - 1}

, which simplifies to

a^2 = 1.20

To find the factor of growth over an interval of 2 units of x.

3

Relate the value of g(6) to the given value of g(2).

g(6) = g(2) \cdot a^4 = g(2) \cdot (a^2)^2

To write the unknown value in terms of the known value and the determined factor

a^2

using exponent rules.

4

Substitute the known values into the equation to calculate the result.

g(6) = 15 \cdot (1.20)^2 = 15 \cdot 1.44 = 21.6

To calculate the final numerical value of g(6).

Key Concept

Exponential Functions and Equations

Estimated Time:2m 30s

Question 38Question →

If $x$ and $y$ are positive numbers with $x > 1$ that satisfy the system of equations below, what is the value of $x$ ?

x^y = 2^{12}

(2x)^{y-2} = 2^8

Show answer & explanation

Answer: 8

Answer

The correct answer is 8.

The correct answer is 8. By manipulating the system of equations, we can express the second equation in terms of

x^y

. Substituting the value

2^{12}

for

x^y

allows us to express

x

in terms of

y

as

x = 2^{(y+2)/2}

. Substituting this back into the first equation results in the quadratic equation

y^2 + 2y - 24 = 0

. Since

y

must be positive, we find

y = 4

. Using this value of

y

, we solve for

x

and obtain

x = 8

, which is greater than 1, satisfying all given conditions.

Step-by-Step Solution

1

Distribute the exponent in the second equation:

(2x)^{y-2} = 2^8

.

2^{y-2} \cdot x^{y-2} = 2^8

Applying the power of a product rule:

(ab)^n = a^n b^n

.

2

Multiply both sides of the equation by

x^2

to align with the term

x^y

.

2^{y-2} \cdot x^y = 2^8 \cdot x^2

Since

x^{y-2} \cdot x^2 = x^{y-2+2} = x^y

, this allows us to substitute the value of

x^y

from the first equation.

3

Substitute

x^y = 2^{12}

into the equation.

2^{y-2} \cdot 2^{12} = 2^8 \cdot x^2

This uses the given system's first equation to eliminate the

x^y

term.

4

Simplify the left side and divide both sides by

2^8

to isolate

x^2

.

x^2 = 2^{y+2}

2^{y-2} \cdot 2^{12} = 2^{y+10}

. Dividing both sides by

2^8

yields

2^{y+10-8} = 2^{y+2}

.

5

Take the square root of both sides to find an expression for

x

.

x = 2^{\frac{y+2}{2}}

Since

x > 1

, we take the positive root:

x = (2^{y+2})^{1/2} = 2^{\frac{y+2}{2}}

.

6

Substitute the expression for

x

back into the first equation

x^y = 2^{12}

.

\left(2^{\frac{y+2}{2}}\right)^y = 2^{12}

This yields a single equation in terms of

y

.

7

Equate the exponents and solve the resulting quadratic equation for

y

.

y = 4

(since

y > 0

)

The equation is

2^{\frac{y^2+2y}{2}} = 2^{12} \implies \frac{y^2+2y}{2} = 12 \implies y^2+2y-24=0 \implies (y+6)(y-4)=0

. Since

y

must be positive,

y = 4

.

8

Substitute

y = 4

back into the expression for

x

.

x = 8

x = 2^{\frac{4+2}{2}} = 2^3 = 8

.

Key Concept

Solving systems of exponential equations by equating bases, applying exponent rules, and solving quadratic equations.

Question 39Question →

A biologist is studying a population of bacteria that triples in size every 4 hours. The population of the bacteria $t$ hours after the start of the study can be modeled by the function $P(t) = P_0 \cdot 3^{kt}$ , where $P_0$ is the initial population and $k$ is a constant. What is the value of $k$ ?

Show answer & explanation

Answer: 0.25

Answer

The correct answer is 0.25 (or 1/4).

The population triples every 4 hours, meaning that after

t

hours, the population has tripled

\frac{t}{4}

times. Thus, the growth model is

P(t) = P_0 \cdot 3^{\frac{t}{4}}

. Comparing this to the given expression

P(t) = P_0 \cdot 3^{kt}

, we get

kt = \frac{t}{4}

. Dividing both sides by

t

yields

k = \frac{1}{4}

, or

0.25

.

Step-by-Step Solution

1

Write the general formula for exponential growth where the population triples every 4 hours.

P(t) = P_0 \cdot 3^{\frac{t}{4}}

If a population triples every 4 hours, it will be multiplied by 3 a total of

\frac{t}{4}

times after

t

hours.

2

Set the exponent of the general model equal to the exponent of the given model.

kt = \frac{t}{4}

Both models represent the same population growth function, so their exponents must be equal.

3

Solve for the constant

k

.

k = \frac{1}{4}

(or

0.25

)

Divide both sides of the equation by

t

.

Key Concept

Exponential growth functions and representing time intervals in the exponent.

Question 40Question →

If $16^{x-1} - 10 \cdot 4^{x-2} + 1 = 0$ , what is the sum of all values of $x$ that satisfy the equation?

Show answer & explanation

Answer: 2

Answer

The sum of all values of

x

that satisfy the equation is 2.

By writing

16^{x-1}

as

\frac{(4^x)^2}{16}

and

10 \cdot 4^{x-2}

as

\frac{10 \cdot 4^x}{16}

, multiplying the entire equation by 16 yields

(4^x)^2 - 10(4^x) + 16 = 0

. Factoring this quadratic equation gives

(4^x - 2)(4^x - 8) = 0

, which yields

4^x = 2

and

4^x = 8

. Solving these exponential equations results in

x = 0.5

and

x = 1.5

. Adding these values together gives

0.5 + 1.5 = 2

.

Step-by-Step Solution

1

Express each exponential term in the equation using a base of 4.

\frac{4^{2x}}{16} - \frac{10 \cdot 4^x}{16} + 1 = 0

Since

16 = 4^2

, we can write

16^{x-1} = (4^2)^{x-1} = 4^{2x-2} = \frac{4^{2x}}{16}

. Similarly,

4^{x-2} = \frac{4^x}{16}

.

2

Multiply every term in the equation by 16 to eliminate the fractions.

4^{2x} - 10 \cdot 4^x + 16 = 0

This simplifies the expression into a standard polynomial form that is quadratic in terms of

4^x

.

3

Use the substitution

u = 4^x

(where

u > 0

) to transform the equation into a quadratic equation in terms of

u

, and solve it.

u^2 - 10u + 16 = 0 \implies (u - 2)(u - 8) = 0 \implies u = 2

or

u = 8

Substituting

u

allows us to factor the quadratic expression using standard factoring techniques.

4

Substitute

4^x

back for

u

and solve for

x

using base 2 conversion.

For

u = 2

:

4^x = 2 \implies 2^{2x} = 2^1 \implies x = 0.5

. For

u = 8

:

4^x = 8 \implies 2^{2x} = 2^3 \implies x = 1.5

.

We must convert the solutions for the intermediate variable

u

back to the original variable

x

.

5

Calculate the sum of all found values of

x

.

0.5 + 1.5 = 2

The question asks for the sum of all real values of

x

that satisfy the equation.

Key Concept

Solving exponential equations of quadratic form by using variable substitution and exponent laws.

Exponential Functions and Equations

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution