Exponential Functions and Equations Practice Questions — SAT — Page 3

Question 41Question →

The population of a colony of bacteria is modeled by the function $P(t) = 250 \cdot 8^{\frac{t}{6} + 1}$ , where $P(t)$ is the estimated number of bacteria $t$ hours after the start of an experiment. The function can be rewritten in the equivalent form $P(t) = A \cdot 2^{\frac{m}{k}}$ , where $m$ is the number of 20-minute intervals that have elapsed since the start of the experiment, and $A$ and $k$ are constants. What is the value of $A + k$ ?

506

2,002

2,006

2,018

Show answer & explanation

Answer: 2,006

Answer

2,006

The correct answer is 2,006. By writing the base 8 as 2^3 and applying the exponent rules, the function becomes 250 * 2^(t/2 + 3) = 2000 * 2^(t/2). Since m represents 20-minute intervals and there are 3 intervals in an hour, we have t = m/3. Substituting this into the function yields 2000 * 2^(m/6), which gives A = 2000 and k = 6. The sum is 2000 + 6 = 2006.

Step-by-Step Solution

1

Express the base 8 in terms of base 2 to match the target form.

P(t) = 250 \cdot (2^3)^{\frac{t}{6} + 1}

Converting to base 2 allows simplification of the exponent to match the required base of 2 in the target equation.

2

Apply the exponent power rule

(b^y)^z = b^{yz}

to simplify the expression.

P(t) = 250 \cdot 2^{3\left(\frac{t}{6} + 1\right)} = 250 \cdot 2^{\frac{t}{2} + 3}

Multiplying the exponents simplifies the expression into a single power of base 2.

3

Isolate the constant coefficient by separating the terms in the exponent using the rule

b^{x+y} = b^x \cdot b^y

.

P(t) = 250 \cdot 2^3 \cdot 2^{\frac{t}{2}} = 250 \cdot 8 \cdot 2^{\frac{t}{2}} = 2000 \cdot 2^{\frac{t}{2}}

This separates the initial coefficient multiplier from the variable growth term.

4

Convert the time variable from hours (

t

) to 20-minute intervals (

m

).

Since there are 3 twenty-minute intervals in one hour,

m = 3t

, which means

t = \frac{m}{3}

. Substituting this into the equation yields

P(t) = 2000 \cdot 2^{\frac{m/3}{2}} = 2000 \cdot 2^{\frac{m}{6}}

.

This expresses the population model in terms of the new time unit variable,

m

, as required.

5

Identify the constants

A

and

k

and calculate their sum.

A = 2000

,

k = 6

, so

A + k = 2000 + 6 = 2006

.

Comparing the final model to

A \cdot 2^{\frac{m}{k}}

directly gives the values of the constants.

Key Concept

Rewriting and interpreting exponential functions by manipulating bases and changing variables.

Estimated Time:2m 0s

Question 42Question →

A researcher models the population of a certain species of bacteria in a culture. The population $N(t)$ of bacteria $t$ hours after the culture is established is modeled by the function

N(t) = N_0 \cdot b^{\frac{t}{3}}

where $N_0$ is the initial population of the bacteria and $b$ is a constant. If the population of the bacteria increases by $44\%$ every 2 hours, what is the value of $b$ ?

1.200

1.275

1.660

1.728

Show answer & explanation

Answer: 1.728

Answer

1.728

The correct answer is

1.728

. Since the population increases by

44\%

every 2 hours, the population at

t = 2

is

1.44

times the initial population, meaning

N(2) = 1.44 \cdot N_0

. Substituting

t = 2

into the model

N(t) = N_0 \cdot b^{\frac{t}{3}}

yields

N_0 \cdot b^{\frac{2}{3}} = 1.44 \cdot N_0

. Dividing both sides by

N_0

gives

b^{\frac{2}{3}} = 1.44

. Raising both sides to the power of

\frac{3}{2}

isolates

b

as

b = (1.44)^{\frac{3}{2}}

. Since

1.44 = (1.2)^2

, we can simplify this expression using exponent rules:

b = (1.2^2)^{\frac{3}{2}} = 1.2^3 = 1.728

.

Step-by-Step Solution

1

Write the equation relating the population at time

t = 2

hours to the initial population at

t = 0

hours using the given percentage increase.

N(2) = 1.44 \cdot N_0

An increase of

44\%

means the population becomes

100\% + 44\% = 144\%

of its initial value, which corresponds to multiplying by a factor of

1.44

.

2

Substitute the function definition

N(t) = N_0 \cdot b^{\frac{t}{3}}

into the equation for

t = 2

.

N_0 \cdot b^{\frac{2}{3}} = 1.44 \cdot N_0

This allows us to set up an equation to solve for the constant

b

.

3

Divide both sides of the equation by

N_0

and isolate

b

.

b^{\frac{2}{3}} = 1.44

Since the initial population

N_0

is positive, we can divide both sides by

N_0

to isolate the exponential base term.

4

Solve for

b

by raising both sides of the equation to the power of

\frac{3}{2}

.

b = (1.44)^{\frac{3}{2}}

To solve for

b

, we multiply the exponent

\frac{2}{3}

by its reciprocal

\frac{3}{2}

.

5

Evaluate the expression

(1.44)^{\frac{3}{2}}

using exponent properties.

b = (1.2^2)^{\frac{3}{2}} = 1.2^3 = 1.728

Expressing

1.44

as

1.2^2

allows us to simplify the fractional exponent using the power of a power rule,

(x^a)^b = x^{ab}

.

Key Concept

Solving exponential equations by applying exponent rules and interpreting exponential growth factors in context.

Question 43Question →

The population of a certain species of plankton in a lake is modeled by the function

P(t) = P_0 \cdot (1.69)^{\frac{t}{d}}

where

P_0

is the initial population,

t

is the time, in days, since the population was first measured, and

d

is a positive constant. The population of the plankton increases by

30\%

every

h

hours, where

h

is a positive constant. If the model is accurate, what is the value of the ratio

\frac{h}{d}

?

0.5

2

12

48

Show answer & explanation

Answer: 12

Answer

The correct value of the ratio is 12.

The correct answer shows the value of 12. Since

t

is measured in days, the time interval of

h

hours must be converted to

\frac{h}{24}

days. A

30\%

increase corresponds to a multiplier of

1.3

. Substituting

t = \frac{h}{24}

into the function gives

(1.69)^{\frac{h}{24d}} = 1.3

. Rewriting

1.69

as

1.3^2

yields

\left(1.3^2\right)^{\frac{h}{24d}} = 1.3

, which simplifies to

1.3^{\frac{h}{12d}} = 1.3^1

. Equating the exponents gives

\frac{h}{12d} = 1

, which solves to

\frac{h}{d} = 12

.

Step-by-Step Solution

1

Express the growth interval in days to match the unit of

t

in the model.

Since there are 24 hours in a day, an interval of

h

hours is equivalent to

\frac{h}{24}

days.

The variable

t

in the model

P(t) = P_0 \cdot (1.69)^{\frac{t}{d}}

represents time in days, so all time variables must be in days.

2

Relate the percentage increase to the population multiplier.

An increase of

30\%

corresponds to a multiplier of

1 + 0.30 = 1.3

.

A

30\%

increase means the new population is

130\%

of the previous population.

3

Set up the equation using the given model at

t = \frac{h}{24}

.

P_0 \cdot (1.69)^{\frac{h}{24d}} = 1.3 \cdot P_0

, which simplifies to

(1.69)^{\frac{h}{24d}} = 1.3

.

At

t = \frac{h}{24}

days, the population must be

1.3

times the initial population

P_0

.

4

Rewrite the base of the exponential term and solve for the ratio.

Since

1.69 = 1.3^2

, the equation becomes

\left(1.3^2\right)^{\frac{h}{24d}} = 1.3^1

, which simplifies to

1.3^{\frac{2h}{24d}} = 1.3^1

, leading to

\frac{h}{12d} = 1

.

Expressing both sides with the same base allows for equating their exponents to solve for the target ratio.

5

Solve for the final ratio

\frac{h}{d}

.

\frac{h}{d} = 12

.

Multiplying both sides of the equation

\frac{h}{12d} = 1

by 12 isolates the ratio

\frac{h}{d}

.

Key Concept

Exponential Equations with Base Manipulation and Unit Conversion

Alternative Method

Instead of substituting a specific value for

t

, we can equate the daily growth factors. The model's daily growth factor is

(1.69)^{1/d}

. A

30\%

increase every

h

hours corresponds to a factor of

1.3

every

h

hours. Since there are

24

hours in a day, there are

\frac{24}{h}

such intervals in a day, yielding a daily growth factor of

(1.3)^{24/h}

. Setting the factors equal:

(1.69)^{1/d} = (1.3)^{24/h} \implies (1.3^2)^{1/d} = (1.3)^{24/h} \implies \frac{2}{d} = \frac{24}{h} \implies \frac{h}{d} = 12

.

Estimated Time:3m 0s

Question 44Question →

The value of a certain investment is modeled by the function $f(t) = 500(1.21)^{\frac{t}{2}}$ , where $f(t)$ represents the value of the investment, in dollars, $t$ years after the initial investment was made. Which of the following is the annual growth rate, as a percentage, of the investment?

21\%

10\%

10.5\%

46.41\%

Show answer & explanation

Answer: $10\%$

Answer

The annual growth rate of the investment is

10\%

.

To find the annual growth rate, we need to rewrite the function

f(t) = 500(1.21)^{\frac{t}{2}}

in the form

f(t) = P(1 + r)^t

, where

P

is the initial value,

1 + r

is the annual growth factor, and

r

is the annual growth rate. Applying the power of a power rule for exponents, we can rewrite the expression as

f(t) = 500\left(1.21^{\frac{1}{2}}\right)^t

. Since raising a number to the power of

\frac{1}{2}

is equivalent to taking the square root, this simplifies to

f(t) = 500(\sqrt{1.21})^t = 500(1.1)^t

. The annual growth factor is

1.1

, which can be written as

1 + 0.1

. Therefore, the annual growth rate

r

is

0.1

, which corresponds to

10\%

.

Step-by-Step Solution

1

Identify the given exponential function and rewrite the expression to isolate the annual growth factor.

The function is

f(t) = 500(1.21)^{\frac{t}{2}}

. Using exponent rules, we can rewrite the term

(1.21)^{\frac{t}{2}}

as

\left(1.21^{\frac{1}{2}}\right)^t

.

Isolating the variable

t

in the exponent allows us to identify the annual growth factor.

2

Calculate the base of the exponential term.

1.21^{\frac{1}{2}} = \sqrt{1.21} = 1.1

. Thus, the rewritten function is

f(t) = 500(1.1)^t

.

Simplifying the base gives us the actual growth factor per unit of time (per year).

3

Determine the percentage growth rate from the growth factor.

The growth factor is

1.1

, which is equal to

1 + 0.1

. The growth rate

r

is

0.1

. Converting

0.1

to a percentage gives

0.1 \times 100\% = 10\%

.

The growth rate is the decimal increase per year, which we convert to a percentage.

Key Concept

Manipulating exponential expressions to identify rates of growth or decay.

Question 45Question →

If $4^x - 3^{x - 0.5} = 3^{x + 0.5} - 2^{2x-1}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1.5

Answer

The correct answer is

1.5

(or

\frac{3}{2}

)

By rewriting

4^x

as

2^{2x}

and grouping the base 2 and base 3 terms, the equation becomes

2^{2x} + 2^{2x-1} = 3^{x+0.5} + 3^{x-0.5}

. Factoring out the variable terms gives

2^{2x}(1 + 2^{-1}) = 3^x(3^{0.5} + 3^{-0.5})

. Simplifying the coefficients results in

\frac{3}{2} \cdot 2^{2x} = \frac{4}{\sqrt{3}} \cdot 3^x

. Dividing both sides to separate the variables and constants yields

\frac{2^{2x}}{3^x} = \frac{8}{3\sqrt{3}}

. Representing both sides with the common base

\frac{4}{3}

gives

\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{1.5}

. Equating the exponents leads to

x = 1.5

.

Step-by-Step Solution

1

Rewrite the base 4 term as a power of 2

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

Expressing exponential terms with prime bases simplifies finding relationships between them.

2

Group terms with like bases on opposite sides of the equation

2^{2x} + 2^{2x-1} = 3^{x+0.5} + 3^{x-0.5}

Separating different base systems allows us to factor out common exponential functions.

3

Factor out the common exponential factors from each side

2^{2x}(1 + 2^{-1}) = 3^x(3^{0.5} + 3^{-0.5})

This isolates the constant coefficients from the exponential functions of

x

.

4

Simplify the constant terms on both sides of the equation

2^{2x}\left(\frac{3}{2}\right) = 3^x\left(\sqrt{3} + \frac{1}{\sqrt{3}}\right) = 3^x\left(\frac{4}{\sqrt{3}}\right)

Reducing the coefficients makes it easier to combine the constant terms.

5

Divide both sides to group the exponential terms together

\frac{2^{2x}}{3^x} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}}

This sets up the equation in the form of

f(x) = C

where

f(x)

has a combined base.

6

Express both sides of the equation with a common base of

\frac{4}{3}

\left(\frac{4}{3}\right)^x = \left(\frac{4}{3}\right)^{1.5}

Since

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

, the left side simplifies to

\left(\frac{4}{3}\right)^x

. The right side can be rewritten as

\frac{8}{3\sqrt{3}} = \frac{2^3}{3^{1.5}} = \frac{(4^{0.5})^3}{3^{1.5}} = \frac{4^{1.5}}{3^{1.5}} = \left(\frac{4}{3}\right)^{1.5}

.

7

Equate the exponents

x = 1.5

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

Key Concept

Solving exponential equations using base conversion, exponent rules, and factoring.

Question 46Question →

If $\frac{8^x}{2^y} = 64$ , which of the following equations correctly expresses $y$ in terms of $x$ ?

y = 3x - 64

y = 3x - 6

y = 3x + 6

y = 6 - 3x

Show answer & explanation

Answer: $y = 3x - 6$

Answer

y = 3x - 6

To solve the equation, express all terms using the common base of 2. Substituting

8 = 2^3

and

64 = 2^6

into the equation gives

\frac{(2^3)^x}{2^y} = 2^6

. Applying the power rule of exponents transforms the numerator into

2^{3x}

, which leads to the equation

\frac{2^{3x}}{2^y} = 2^6

. Using the quotient rule of exponents, the division simplifies to

2^{3x - y} = 2^6

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

3x - y = 6

. Isolating the variable yields the correct formulation.

Step-by-Step Solution

1

Rewrite all bases in the equation as powers of the common base 2.

Since

8 = 2^3

and

64 = 2^6

, the equation becomes

\frac{(2^3)^x}{2^y} = 2^6

.

Expressing all exponential terms using a single base allows for the application of exponent rules.

2

Apply the power of a power property to simplify the numerator on the left side.

The term

(2^3)^x

simplifies to

2^{3x}

, yielding the equation

\frac{2^{3x}}{2^y} = 2^6

.

The exponent rule

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

states that a base raised to a power and then another power requires multiplying the exponents.

3

Apply the quotient property of exponents to combine the left side of the equation.

The left side simplifies to

2^{3x - y}

, resulting in

2^{3x - y} = 2^6

.

The quotient rule

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

states that dividing powers with the same base requires subtracting the exponent of the denominator from the exponent of the numerator.

4

Set the exponents equal to each other and solve for the desired variable.

3x - y = 6

, which simplifies to

y = 3x - 6

when solved for

y

.

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

Key Concept

Solving exponential equations by base conversion and rules of exponents

Alternative Method

Instead of converting both sides to base 2, you can also substitute a test value for

x

. For example, if

x = 2

, then the equation becomes

\frac{8^2}{2^y} = 64

, which simplifies to

\frac{64}{2^y} = 64 \implies 2^y = 1 \implies y = 0

. Substituting

x = 2

into the correct option yields

y = 3(2) - 6 = 0

, confirming the answer.

Estimated Time:1m 30s

Question 47Question →

The concentration of a medication in a patient's bloodstream, in milligrams per liter ( $\text{mg/L}$ ), $h$ hours after injection is modeled by an exponential decay function. If the concentration of the medication decreases by $48.8\%$ every $3$ hours, which of the following is closest to the percent decrease of the medication concentration each hour?

16.3%

20.0%

80.0%

86.6%

Show answer & explanation

Answer: 20.0%

Answer

20.0%

The correct answer is 20.0%. Since the concentration decreases by 48.8% every 3 hours, the remaining concentration after 3 hours is 51.2%, which corresponds to a factor of 0.512 of the initial concentration. The hourly decay factor

b

satisfies the equation

b^3 = 0.512

. Taking the cube root of both sides yields

b = 0.8

. A decay factor of 0.8 means that 80% of the medication remains each hour, representing a percent decrease of 20% each hour.

Step-by-Step Solution

1

Determine the remaining fraction of the medication after 3 hours.

Remaining fraction = 1 - 0.488 = 0.512

Since the concentration decreases by 48.8% every 3 hours, the remaining concentration is 51.2% of its value at the beginning of the 3-hour interval.

2

Set up the equation for the hourly retention factor,

b

.

b^3 = 0.512

An exponential decay model of the form

C(h) = C_0 \cdot b^h

retains a factor of

b^3

over a 3-hour period.

3

Solve for the hourly retention factor,

b

.

b = \sqrt[3]{0.512} = 0.8

Taking the cube root of 0.512 gives the hourly factor, indicating that the concentration retains 80% of its value each hour.

4

Calculate the hourly percent decrease from the hourly retention factor.

Percent decrease =

(1 - 0.8) \times 100\% = 20.0\%

The hourly percent decrease is the difference between the initial 100% and the remaining 80% retained each hour.

Key Concept

Converting decay factors across different time intervals using exponent properties.

Estimated Time:2m 0s

Question 48Question →

If $x$ and $y$ satisfy the system of equations below, what is the value of $x + y$ ?

\begin{aligned} 8^{x-y} &= 2^{y+3} \\ 9^{x} &= 27^{y-1} \end{aligned}

Show answer & explanation

Answer: 36

Answer

36

The correct answer is 36. By rewriting

8^{x-y} = 2^{y+3}

with base 2, we obtain the relation

3x - 4y = 3

. Similarly, rewriting

9^x = 27^{y-1}

with base 3 gives

2x - 3y = -3

. Solving this system of equations simultaneously yields

x = 21

and

y = 15

. Adding these values together gives the final result of 36.

Step-by-Step Solution

1

Rewrite the first equation using the common base of 2.

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

, which simplifies to

3x - 4y = 3

.

Since

8 = 2^3

, we can apply the power of a power rule

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

to rewrite

8^{x-y}

as

2^{3(x-y)}

. Setting the exponents equal gives the linear equation

3(x-y) = y+3

, which simplifies to

3x - 4y = 3

.

2

Rewrite the second equation using the common base of 3.

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

, which simplifies to

2x - 3y = -3

.

Since

9 = 3^2

and

27 = 3^3

, we rewrite the terms using base 3. Equating the exponents gives

2x = 3(y-1)

, which simplifies to

2x - 3y = -3

.

3

Solve the system of two linear equations.

x = 21

and

y = 15

.

Using the elimination method, we can multiply the first equation by 3 and the second equation by 4 to align the coefficients of

y

, obtaining

9x - 12y = 9

and

8x - 12y = -12

. Subtracting the equations gives

x = 21

. Substituting this back yields

y = 15

.

4

Find the sum of

x

and

y

.

36

The question asks for the value of

x + y

. Substituting

x = 21

and

y = 15

gives

21 + 15 = 36

.

Key Concept

Solving systems of exponential equations by expressing bases as powers of common bases to establish a system of linear equations.

Question 49Question →

Two different cultures of bacteria, culture $A$ and culture $B$ , begin growing at the same time. The population of culture $A$ increases by $300\%$ every $8$ hours, and the population of culture $B$ increases by $700\%$ every $9$ hours. If both cultures start with the same initial population, after how many hours will the population of culture $B$ be $4$ times the population of culture $A$ ?

9.6

24

2

48

Show answer & explanation

Answer: 24

Answer

The population of culture B will be 4 times the population of culture A after 24 hours.

The correct answer is the option stating 24 hours. The growth model for culture A is

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

and for culture B is

P_B(t) = P_0 \cdot 8^{t/9}

. Setting

P_B(t) = 4 P_A(t)

and converting all terms to base 2 yields

2^{t/3} = 2^{2 + t/4}

. Equating exponents gives

t/3 = 2 + t/4

, which simplifies to

t/12 = 2

and solves to

t = 24

hours.

Step-by-Step Solution

1

Determine the growth factors and set up the population equations for both cultures.

For culture

A

, a

300\%

increase means the population becomes

1 + 3.00 = 4

times its previous value every

8

hours, so

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

. For culture

B

, a

700\%

increase means the population becomes

1 + 7.00 = 8

times its previous value every

9

hours, so

P_B(t) = P_0 \cdot 8^{t/9}

.

Establishing the correct exponential growth models with the appropriate growth factors and time constants is necessary to relate the populations.

2

Set up the equation to find when the population of culture

B

is

4

times that of culture

A

.

P_B(t) = 4 P_A(t) \implies P_0 \cdot 8^{t/9} = 4 \cdot P_0 \cdot 4^{t/8} \implies 8^{t/9} = 4 \cdot 4^{t/8}

This sets up the equation that must be solved for

t

by dividing both sides by the non-zero initial population

P_0

.

3

Express all terms with a common base of

2

.

(2^3)^{t/9} = 2^2 \cdot (2^2)^{t/8} \implies 2^{t/3} = 2^2 \cdot 2^{t/4}

Expressing bases as powers of

2

allows the use of exponent rules to simplify the equation.

4

Apply the product rule of exponents to combine the terms on the right side.

2^{t/3} = 2^{2 + t/4}

The rule

2^a \cdot 2^b = 2^{a+b}

simplifies the multiplication of exponential terms with the same base.

5

Equate the exponents and solve the resulting linear equation for

t

.

t/3 = 2 + t/4 \implies t/3 - t/4 = 2 \implies t/12 = 2 \implies t = 24

Since the bases are equal, their exponents must be equal. Solving the linear equation yields the time in hours.

Key Concept

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

Estimated Time:3m 0s

Question 50Question →

The table below shows the population of a colony of bacteria, $P(t)$ , at various times $t$ , in hours.

Time $t$ (hours)	Population $P(t)$
0	120
3	960
6	7,680

If the population can be modeled by an exponential function of the form $P(t) = a \cdot b^t$ , where $a$ and $b$ are constants, what is the value of $b$ ?

2

8

4

\frac{8}{3}

Show answer & explanation

Answer: 2

Answer

2

The correct answer is the value that represents the hourly growth factor of the bacteria. Since the initial population at

t = 0

is 120, we have

a = 120

. Using the population at

t = 3

, we set up the equation

120 \cdot b^3 = 960

. Dividing both sides by 120 yields

b^3 = 8

. Taking the cube root of 8 gives

b = 2

. We can verify this with the third data point:

120 \cdot 2^6 = 120 \cdot 64 = 7,680

, which matches the table.

Step-by-Step Solution

1

Find the value of the constant

a

using the initial value from the table.

a = 120

When

t = 0

, the population

P(0) = 120

. Substituting these values into

P(t) = a \cdot b^t

gives

120 = a \cdot b^0

, and since

b^0 = 1

, we get

a = 120

.

2

Set up an equation to solve for

b

using another data point from the table.

120 \cdot b^3 = 960

When

t = 3

, the population

P(3) = 960

. Substituting

a = 120

,

t = 3

, and

P(3) = 960

into the function gives

960 = 120 \cdot b^3

.

3

Isolate

b^3

and solve for

b

by taking the cube root.

b = 2

Dividing both sides of

120 \cdot b^3 = 960

by

120

gives

b^3 = 8

. Taking the cube root of both sides gives

b = \sqrt[3]{8} = 2

.

Key Concept

Determining parameters of exponential functions from a table of values.

Question 51Question →

If $\frac{\left(\frac{1}{4}\right)^{2x-5}}{8^{3-x}} = \left(\sqrt[3]{2}\right)^{x-5}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

2

By writing all terms with a common base of 2, the equation simplifies to

2^{-x+1} = 2^{\frac{x-5}{3}}

. Equating the exponents yields

-x + 1 = \frac{x-5}{3}

. Solving this linear equation gives

x = 2

.

Step-by-Step Solution

1

Rewrite each base in terms of base 2.

\frac{1}{4} = 2^{-2}

,

8 = 2^3

, and

\sqrt[3]{2} = 2^{\frac{1}{3}}

Expressing all terms with a common base allows the exponents to be directly compared.

2

Substitute the bases back into the left side of the equation and simplify using exponent rules.

\frac{(2^{-2})^{2x-5}}{(2^3)^{3-x}} = \frac{2^{-4x+10}}{2^{9-3x}} = 2^{-x+1}

Applying the power rule

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

and quotient rule

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

simplifies the fraction.

3

Rewrite the right side of the equation using base 2.

(\sqrt[3]{2})^{x-5} = (2^{\frac{1}{3}})^{x-5} = 2^{\frac{x-5}{3}}

Applying the power rule to the fractional exponent converts the radical expression.

4

Equate the simplified left and right sides, then set the exponents equal to each other.

-x + 1 = \frac{x-5}{3}

If two exponential expressions with the same base are equal, their exponents must be equal.

5

Solve the linear equation for

x

.

-3x + 3 = x - 5 \implies 8 = 4x \implies x = 2

Isolating the variable

x

yields the final solution.

Key Concept

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

Question 52Question →

The mass of a radioactive sample, in grams, is modeled by the function $M(t) = M_0 \left(\frac{1}{8}\right)^{\frac{t}{15}}$ , where $M_0$ is the initial mass of the sample and $t$ is the time, in minutes, since the sample started decaying. Which of the following functions correctly models the mass of the sample, in grams, in terms of $s$ seconds after the sample started decaying?

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{5}}

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{2700}}

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

M(s) = M_0 \left(\frac{1}{2}\right)^{12s}

Show answer & explanation

Answer: $M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}$

Answer

The correct function modeling the mass in terms of seconds is

M(s) = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

.

To express the mass in terms of

s

seconds, we substitute

t = \frac{s}{60}

into the original decay function because the time in minutes is equal to the number of seconds divided by 60. This gives

M(s) = M_0 \left(\frac{1}{8}\right)^{\frac{s}{60 \times 15}} = M_0 \left(\frac{1}{8}\right)^{\frac{s}{900}}

. Next, since the options have a base of

\frac{1}{2}

, we rewrite the base

\frac{1}{8}

as

\left(\frac{1}{2}\right)^3

. Applying the power of a power rule

(x^a)^b = x^{ab}

, we multiply the exponent

\frac{s}{900}

by 3, which yields

M(s) = M_0 \left(\frac{1}{2}\right)^{3 \times \frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

.

Step-by-Step Solution

1

Relate the time variables

t

(in minutes) and

s

(in seconds).

Since 1 minute is equivalent to 60 seconds, the relationship is

t = \frac{s}{60}

.

This substitution allows the function to take time input in seconds instead of minutes.

2

Substitute

t = \frac{s}{60}

into the exponent of the original model.

M(s) = M_0 \left(\frac{1}{8}\right)^{\frac{s/60}{15}} = M_0 \left(\frac{1}{8}\right)^{\frac{s}{900}}

This updates the function variable to seconds and simplifies the fractional exponent.

3

Rewrite the base

\frac{1}{8}

as a power of

\frac{1}{2}

and apply exponent rules.

Since

\frac{1}{8} = \left(\frac{1}{2}\right)^3

, we write

M(s) = M_0 \left(\left(\frac{1}{2}\right)^3\right)^{\frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{3 \times \frac{s}{900}} = M_0 \left(\frac{1}{2}\right)^{\frac{s}{300}}

.

Applying the power of a power rule,

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

, simplifies the function to its final form with a base of

\frac{1}{2}

.

Key Concept

Applying exponent rules to manipulate exponential bases and performing variable substitutions in contextual models.

Estimated Time:2m 0s

Question 53Question →

If $81^{y-2} = 27^{y+1}$ , what is the value of $y$ ?

Show answer & explanation

Answer: 11

Answer

11

The correct answer is 11. By expressing both sides of the equation with the common base of 3, the equation simplifies from

81^{y-2} = 27^{y+1}

to

(3^4)^{y-2} = (3^3)^{y+1}

. Applying the power of a power rule gives

3^{4y-8} = 3^{3y+3}

. Equating the exponents yields

4y - 8 = 3y + 3

, which solves to

y = 11

.

Step-by-Step Solution

1

Express both bases as powers of 3

(3^4)^{y-2} = (3^3)^{y+1}

To solve an exponential equation algebraically, it is helpful to express both sides using a common base.

2

Apply the power of a power exponent rule

3^{4(y-2)} = 3^{3(y+1)}

The rule

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

allows us to simplify the exponent expressions by multiplying the exponents.

3

Equate the exponents

4(y-2) = 3(y+1)

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

4

Distribute the coefficients

4y - 8 = 3y + 3

Expanding the linear expressions prepares the equation for isolation of the variable.

5

Solve the linear equation for

y

y = 11

Subtract

3y

from both sides to get

y - 8 = 3

, then add 8 to both sides to isolate

y

.

Key Concept

Solving exponential equations by expressing bases in terms of a common base and equating exponents.

Alternative Method

Alternatively, substitute the value of 11 back into the original equation to verify that both sides are equal:

81^{11-2} = 81^9 = (3^4)^9 = 3^{36}

and

27^{11+1} = 27^{12} = (3^3)^{12} = 3^{36}

.

Estimated Time:1m 30s

Question 54Question →

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

3

7

8

9

Show answer & explanation

Answer: 7

Answer

The correct value of

x

is 7.

To solve the equation

9^{2x - 1} = \frac{27^{x + 2}}{3}

, write all bases as powers of 3:

9 = 3^2

and

27 = 3^3

. Substituting these into the equation gives

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

. Applying the power rule of exponents, we get

3^{4x - 2} = \frac{3^{3x + 6}}{3^1}

. Using the quotient rule of exponents on the right side, we subtract the exponent in the denominator (which is 1) from the exponent in the numerator:

3^{3x + 6 - 1} = 3^{3x + 5}

. Now we have

3^{4x - 2} = 3^{3x + 5}

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

4x - 2 = 3x + 5

. Solving for

x

by subtracting

3x

and adding 2 to both sides gives the correct value of 7.

Step-by-Step Solution

1

Express all bases as powers of 3.

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

To solve exponential equations with different bases, we rewrite them using a common base to apply exponent laws.

2

Apply the power of a power rule to simplify exponents.

3^{4x - 2} = \frac{3^{3x + 6}}{3^1}

Multiply exponents when raising a power to another power:

(a^m)^n = a^{m \cdot n}

.

3

Apply the quotient rule to simplify the right side of the equation.

3^{4x - 2} = 3^{3x + 5}

Subtract the exponent of the denominator from the numerator when dividing:

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

.

4

Set the exponents equal to each other and solve the linear equation.

x = 7

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

4x - 2 = 3x + 5

.

Key Concept

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

Question 55Question →

An environmental study monitors the populations of two different plant species in a conservation area. The population of Species A is modeled by a linear function, $A(t) = 120 + 15t$ , where $t$ represents the number of years since the start of the study. The population of Species B is modeled by an exponential function, $B(t) = c \cdot d^t$ , where $c$ and $d$ are constants. At the start of the study ( $t = 0$ ), the population of Species A is $4$ times the population of Species B. After $2$ years, the population of Species A is equal to the population of Species B. What is the population of Species B after $4$ years?

Show answer & explanation

Answer: 750

Answer

The population of Species B after

4

years is

750

.

To find the population of Species B after

4

years, we evaluate the models at the given points. At

t = 0

,

A(0) = 120

. Since the population of Species A is

4

times that of Species B at

t = 0

, the initial population of Species B is

30

, which gives

c = 30

. At

t = 2

,

A(2) = 120 + 15(2) = 150

. Since the populations are equal at

t = 2

, we have

B(2) = 30 \cdot d^2 = 150

, which simplifies to

d^2 = 5

. The population of Species B at

t = 4

is given by

B(4) = 30 \cdot d^4 = 30 \cdot (d^2)^2 = 30 \cdot 5^2 = 750

.

Step-by-Step Solution

1

Find the population of Species A at the start of the study (

t = 0

)

A(0) = 120

This establishes the baseline population of Species A to find the corresponding initial population of Species B.

2

Determine the constant

c

, which represents the initial population of Species B

c = 30

Since the population of Species A is

4

times that of Species B at

t = 0

, we solve

120 = 4c

.

3

Calculate the population of Species A after

2

years (

t = 2

)

A(2) = 150

This value is needed because the population of Species B equals the population of Species A at

t = 2

.

4

Solve for the growth factor term

d^2

d^2 = 5

Using the equality

B(2) = 150

, we solve

30 \cdot d^2 = 150

.

5

Calculate the population of Species B after

4

years (

t = 4

)

B(4) = 750

Using the exponential model

B(t) = 30 \cdot d^t

, we find

B(4) = 30 \cdot d^4 = 30 \cdot (d^2)^2 = 30 \cdot 5^2

.

Key Concept

Solving systems involving linear and exponential models using initial conditions and key points.

Question 56Question →

If $\frac{16^{x+1}}{2^{x-3}} = 64$ , what is the value of $x$ ?

\frac{2}{3}

\frac{5}{3}

-\frac{1}{3}

-\frac{7}{3}

Show answer & explanation

Answer: $-\frac{1}{3}$

Answer

-\frac{1}{3}

The correct answer is

-\frac{1}{3}

. By expressing all terms with base 2, the equation

\frac{16^{x+1}}{2^{x-3}} = 64

becomes

\frac{2^{4x+4}}{2^{x-3}} = 2^6

. Applying exponent division rules yields

2^{(4x+4)-(x-3)} = 2^6

, which simplifies to

2^{3x+7} = 2^6

. Setting the exponents equal gives

3x + 7 = 6

, which simplifies to

x = -\frac{1}{3}

.

Step-by-Step Solution

1

Rewrite all terms in the equation using a common base of 2.

The equation becomes

\frac{(2^4)^{x+1}}{2^{x-3}} = 2^6

.

Expressing bases 16 and 64 as powers of 2 allows us to apply exponent laws to simplify the equation.

2

Apply the power of a power rule to the numerator, then apply the quotient rule for exponents.

The left side simplifies to

2^{4x+4 - (x-3)} = 2^{3x+7}

.

The expression

(2^4)^{x+1}

becomes

2^{4x+4}

. Using the quotient rule

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

, we subtract the denominator's exponent

(x-3)

from the numerator's exponent

(4x+4)

, distributing the negative sign to get

3x+7

.

3

Set the exponents equal to each other and solve the resulting linear equation.

3x + 7 = 6 \implies 3x = -1 \implies x = -\frac{1}{3}

.

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

Key Concept

Solving exponential equations by rewriting with a common base

Estimated Time:1m 30s

Question 57Question →

If $27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

To solve the equation

27^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}

, express both sides with the base

3

. Since

27 = 3^3

and

\frac{1}{3} = 3^{-1}

, the equation becomes

(3^3)^{2x - 2} = (3^{-1})^{x - 8}

. Applying the exponent rule

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

yields

3^{6x - 6} = 3^{8 - x}

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

6x - 6 = 8 - x

. Adding

x

to both sides gives

7x - 6 = 8

, and adding

6

to both sides gives

7x = 14

. Dividing by

7

results in

x = 2

.

Step-by-Step Solution

1

Rewrite

27

and

\frac{1}{3}

as powers of

3

.

27 = 3^3

and

\frac{1}{3} = 3^{-1}

This allows both sides of the equation to be expressed with a common base of

3

so their exponents can be equated.

2

Substitute these bases back into the equation and apply exponent rules.

(3^3)^{2x - 2} = (3^{-1})^{x - 8} \implies 3^{3(2x - 2)} = 3^{-(x - 8)} \implies 3^{6x - 6} = 3^{8 - x}

Using the power rule

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

, multiply the outer exponent by the power of the base.

3

Equate the exponents and solve for

x

.

6x - 6 = 8 - x \implies 7x = 14 \implies x = 2

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

Key Concept

Solving exponential equations by finding a common base and applying exponent rules.

Question 58Question →

If $4^{a} \cdot 8^{b} = 32^{5}$ and $a + b = 9$ , what is the value of $b$ ?

2

5

7

11

Show answer & explanation

Answer: 7

Answer

7

By converting all bases to 2, the equation

4^a \cdot 8^b = 32^5

becomes

(2^2)^a \cdot (2^3)^b = (2^5)^5

, which simplifies to

2^{2a+3b} = 2^{25}

. Equating the exponents gives the linear equation

2a + 3b = 25

. Since we are given that

a + b = 9

, we can multiply this equation by 2 to get

2a + 2b = 18

. Subtracting the two equations yields

b = 7

.

Step-by-Step Solution

1

Express each base in the equation

4^a \cdot 8^b = 32^5

as a power of 2.

Since

4 = 2^2

,

8 = 2^3

, and

32 = 2^5

, the equation becomes

(2^2)^a \cdot (2^3)^b = (2^5)^5

.

To solve exponential equations with different bases, it is helpful to express all terms using a common base.

2

Apply the power-of-a-power property

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

and the product-of-powers property

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

to simplify both sides.

The left side simplifies to

2^{2a} \cdot 2^{3b} = 2^{2a+3b}

, and the right side simplifies to

2^{25}

. The equation is now

2^{2a+3b} = 2^{25}

.

Simplifying the expressions allows us to equate the exponents.

3

Set the exponents equal to each other to form a linear equation.

2a + 3b = 25

.

If two exponential expressions with the same base are equal, their exponents must be equal.

4

Solve the system of equations consisting of

2a + 3b = 25

and

a + b = 9

.

Multiplying the second equation by 2 gives

2a + 2b = 18

. Subtracting this from

2a + 3b = 25

yields

(2a + 3b) - (2a + 2b) = 25 - 18

, which simplifies to

b = 7

.

Eliminating one variable allows us to solve for the other variable directly.

Key Concept

Solving exponential equations by converting terms to a common base and solving the resulting system of linear equations.

Alternative Method

Substitute the answer options for

b

back into the equations. If

b = 7

, then

a = 9 - 7 = 2

. Plugging these values into the left side of the exponential equation gives

4^2 \cdot 8^7 = 16 \cdot (2^3)^7 = 2^4 \cdot 2^{21} = 2^{25}

. The right side is

32^5 = (2^5)^5 = 2^{25}

. Since both sides are equal, 7 is the correct answer.

Estimated Time:1m 30s

Question 59Question →

If $\frac{4^{x+3}}{8^{x-1}} = 16^{3-x}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1

Answer

1

The correct answer is 1. By writing all terms with a base of 2, the equation is expressed as

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

. Applying the power rule to simplify the exponents yields

\frac{2^{2x+6}}{2^{3x-3}} = 2^{12-4x}

. Using the quotient rule, we subtract the exponents on the left-hand side to get

2^{(2x+6)-(3x-3)} = 2^{-x+9}

. Equating the exponents gives the linear equation

-x+9 = 12-4x

. Adding

4x

and subtracting

9

from both sides results in

3x = 3

, which simplifies to

x=1

.

Step-by-Step Solution

1

Express all exponential terms using a common base of 2

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

Rewriting each base as a power of 2 allows the application of standard exponent rules.

2

Apply the power of a power rule

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

to distribute the exponents

\frac{2^{2x+6}}{2^{3x-3}} = 2^{12-4x}

Multiplying the inner exponent by each term of the outer exponent simplifies the expression.

3

Apply the quotient of powers rule

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

to combine the fraction

2^{-x+9} = 2^{12-4x}

Subtracting the exponent in the denominator from the exponent in the numerator simplifies the left-hand side.

4

Equate the exponents since the bases are equal

-x + 9 = 12 - 4x

If two exponential expressions with the same positive base are equal, their exponents must be equal.

5

Solve the linear equation for

x

x = 1

Isolating

x

by algebraic manipulation yields the final answer.

Key Concept

Solving exponential equations by finding a common base and applying the laws of exponents.

Question 60Question →

A sample of a radioactive isotope decays according to the function $A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$ , where $A_0$ is the initial mass of the sample, $t$ is the time in years, and $h$ is the half-life of the isotope in years. If the sample decays to $12.5\%$ of its initial mass in $45$ years, what is the value of $h$ ?

5.625

22.5

15

135

Show answer & explanation

Answer: 15

Answer

15

The correct answer is 15. The remaining fraction of the radioactive isotope is

12.5\%

, which can be written as

0.125

or

\frac{1}{8}

. Expressing this as a power of the decay base yields

\frac{1}{8} = \left(\frac{1}{2}\right)^3

. Setting the decay formula equal to this fraction gives

\left(\frac{1}{2}\right)^{\frac{45}{h}} = \left(\frac{1}{2}\right)^3

. Equating the exponents gives

\frac{45}{h} = 3

, which yields

h = 15

.

Step-by-Step Solution

1

Set up the decay equation based on the given remaining percentage.

A(45) = 0.125 A_0

, which simplifies to

\frac{A(45)}{A_0} = 0.125 = \frac{1}{8}

.

This establishes the fraction of the substance remaining after

45

years.

2

Substitute the remaining fraction and the time

t = 45

into the exponential decay model.

\left(\frac{1}{2}\right)^{\frac{45}{h}} = \frac{1}{8}

.

This sets up the equation in terms of the unknown half-life parameter

h

.

3

Express both sides of the equation with a common base of

\frac{1}{2}

.

\left(\frac{1}{2}\right)^{\frac{45}{h}} = \left(\frac{1}{2}\right)^3

.

Converting

\frac{1}{8}

to

\left(\frac{1}{2}\right)^3

allows us to equate the exponents directly.

4

Equate the exponents and solve for

h

.

\frac{45}{h} = 3 \implies 3h = 45 \implies h = 15

.

Since the bases are identical, their exponents must be equal to satisfy the equation.

Key Concept

Solving exponential equations by expressing both sides with a common base.

Estimated Time:1m 30s

Exponential Functions and Equations

Answer

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