Logarithmic and Exponential Expressions and Equations Practice Questions — ACT

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

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

Rewriting

27

as

3^3

gives the equation

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

. Equating the exponents yields

2x - 1 = 3

, which solves to

x = 2

.

Step-by-Step Solution

1

Rewrite the right side of the equation with a base of

3

.

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

Expressing both sides of the equation with a common base allows for direct comparison of the exponents.

2

Set the exponents equal to each other.

2x - 1 = 3

Since the bases are both

3

, their exponents must be equal for the expressions to be equal.

3

Solve the linear equation for

x

.

x = 2

Adding

1

to both sides gives

2x = 4

. Dividing both sides by

2

results in

x = 2

.

Key Concept

Solving exponential equations by expressing both sides with a common base

Question 2Question →

If $\log_5 x = 3$ , what is the value of $x$ ?

8

15

25

125

243

Show answer & explanation

Answer: 125

Answer

125

To solve the logarithmic equation

\log_5 x = 3

, we apply the fundamental definition of a logarithm. A logarithmic equation of the form

\log_b x = y

can be rewritten in exponential form as

b^y = x

. In this equation, the base

b

is 5 and the exponent

y

is 3. Rewriting gives

5^3 = x

. Evaluating

5^3

yields

5 \times 5 \times 5 = 125

. Therefore, the correct value of

x

is 125.

Step-by-Step Solution

1

Apply the definition of a logarithm to rewrite the logarithmic equation in its equivalent exponential form.

5^3 = x

By definition,

\log_b x = y

is equivalent to

b^y = x

, where

b

is the base,

y

is the exponent, and

x

is the argument.

2

Evaluate the exponential expression

5^3

to solve for

x

.

x = 125

Cubing 5 means multiplying it by itself three times:

5 \times 5 \times 5 = 125

.

Key Concept

Definition of Logarithms

Question 3Question →

If $x$ and $y$ are positive real numbers greater than $1$ such that $\log_y(x) + 6\log_x(y) = 5$ and $xy = 64$ , what is the sum of all possible values of $x$ ?

16 + 16\sqrt{2}

4 + 2\sqrt{2}

4\sqrt{2}

80

528

Show answer & explanation

Answer: $16 + 16\sqrt{2}$

Answer

16 + 16\sqrt{2}

Using the change of base property

\log_x(y) = \frac{1}{\log_y(x)}

, the equation can be written in terms of

u = \log_y(x)

as

u + \frac{6}{u} = 5

. Solving this quadratic equation gives

u = 2

or

u = 3

. If

\log_y(x) = 2

, then

x = y^2

. Substituting into

xy = 64

yields

y^3 = 64 \implies y = 4

, which gives

x = 16

. If

\log_y(x) = 3

, then

x = y^3

. Substituting into

xy = 64

yields

y^4 = 64 \implies y = 2\sqrt{2}

, which gives

x = 16\sqrt{2}

. The sum of these values is

16 + 16\sqrt{2}

.

Step-by-Step Solution

1

Apply the change of base formula to express the equation in terms of a single logarithmic base.

Since

\log_x(y) = \frac{1}{\log_y(x)}

, the equation

\log_y(x) + 6\log_x(y) = 5

becomes

\log_y(x) + \frac{6}{\log_y(x)} = 5

.

This allows us to solve for the log expression using substitution.

2

Substitute

u = \log_y(x)

and solve the resulting quadratic equation for

u

.

u + \frac{6}{u} = 5 \implies u^2 - 5u + 6 = 0 \implies (u-2)(u-3) = 0

, so

u = 2

or

u = 3

.

Solving the quadratic equation gives the possible relationships between

x

and

y

.

3

Analyze the first case where

u = 2

and solve for

x

.

If

\log_y(x) = 2

, then

x = y^2

. Substituting this into

xy = 64

gives

y^3 = 64

, which yields

y = 4

. Thus,

x = 4^2 = 16

.

This determines the first possible value of

x

.

4

Analyze the second case where

u = 3

and solve for

x

.

If

\log_y(x) = 3

, then

x = y^3

. Substituting this into

xy = 64

gives

y^4 = 64

, which yields

y = 64^{1/4} = (2^6)^{1/4} = 2^{3/2} = 2\sqrt{2}

. Thus,

x = (2\sqrt{2})^3 = 16\sqrt{2}

.

This determines the second possible value of

x

.

5

Sum the possible values of

x

.

16 + 16\sqrt{2}

The question asks for the sum of all possible values of

x

.

Key Concept

Solving systems of exponential and logarithmic equations using base-change properties and substitution

Alternative Method

Instead of using substitution directly, you can write both equations in terms of base 2 or natural logs: let

\log_y(x) = k

, which means

x = y^k

. We then have

k + 6/k = 5

giving

k = 2

or

k = 3

. This leads directly to

x = y^2

and

x = y^3

, which can then be substituted into the second equation.

Estimated Time:3m 0s

Question 4Question →

If $x$ is a positive real number such that $\log_2(x) + \log_4(x) + \log_{16}(x) = 7$ , what is the value of $x$ ?

Show answer & explanation

Answer: 16

Answer

The value of

x

is

16

.

By converting all logarithms to base 2, we write

\log_4(x)

as

\frac{1}{2}\log_2(x)

and

\log_{16}(x)

as

\frac{1}{4}\log_2(x)

. Summing these gives

\frac{7}{4}\log_2(x) = 7

, which simplifies to

\log_2(x) = 4

. Converting to exponential form, we find

x = 2^4 = 16

.

Step-by-Step Solution

1

Apply the change of base formula to express the logarithms with a common base of 2.

\log_4(x) = \frac{1}{2}\log_2(x)

and

\log_{16}(x) = \frac{1}{4}\log_2(x)

, giving the equation

\log_2(x) + \frac{1}{2}\log_2(x) + \frac{1}{4}\log_2(x) = 7

.

Rewriting the terms with a common base allows them to be combined algebraically.

2

Combine the coefficients of the like terms on the left-hand side.

\frac{7}{4}\log_2(x) = 7

.

The sum of the coefficients is

1 + \frac{1}{2} + \frac{1}{4} = \frac{4}{4} + \frac{2}{4} + \frac{1}{4} = \frac{7}{4}

.

3

Isolate the logarithm term by dividing or multiplying by the reciprocal coefficient.

\log_2(x) = 4

.

Multiplying both sides by

\frac{4}{7}

solves for the value of

\log_2(x)

.

4

Convert the equation from logarithmic form to its equivalent exponential form.

x = 2^4 = 16

.

By definition,

\log_b(a) = c

is equivalent to

b^c = a

.

Key Concept

Change of Base Formula for Logarithms

Question 5Question →

If $\log_2(a) + \log_2(b) = 5$ and $\log_2(a^2) - \log_2(b) = 4$ for positive real numbers $a$ and $b$ , what is the value of $a + b$ ?

5

10

12

16

32

Show answer & explanation

Answer: 12

Answer

12

The correct answer is 12. We can simplify the system of equations by using the power rule of logarithms, which allows us to rewrite

\log_2(a^2)

as

2\log_2(a)

. Letting

x = \log_2(a)

and

y = \log_2(b)

gives us the system

x + y = 5

and

2x - y = 4

. Adding these equations gives

3x = 9

, which means

x = 3

. Substituting this back gives

y = 2

. Converting back from logarithmic form to exponential form, we get

a = 2^3 = 8

and

b = 2^2 = 4

. Therefore,

a + b = 8 + 4 = 12

.

Step-by-Step Solution

1

Use the power property of logarithms,

\log_b(x^k) = k \log_b(x)

, to rewrite the second equation.

The equation

\log_2(a^2) - \log_2(b) = 4

becomes

2\log_2(a) - \log_2(b) = 4

.

This simplifies the term

\log_2(a^2)

so that it is linear in terms of

\log_2(a)

.

2

Substitute variables to simplify solving the system of equations. Let

x = \log_2(a)

and

y = \log_2(b)

.

The system of equations becomes:
1)

x + y = 5

2)

2x - y = 4

Variable substitution reduces the logarithmic system to a standard system of linear equations.

3

Solve the linear system by adding the two equations together.

Adding the equations yields

(x + y) + (2x - y) = 5 + 4

, which simplifies to

3x = 9

, so

x = 3

. Substituting

x = 3

back into the first equation gives

3 + y = 5

, so

y = 2

.

Addition eliminates the variable

y

, allowing us to solve for

x

and then find

y

.

4

Convert the solved values of

x

and

y

back into

a

and

b

using the exponential form definition of a logarithm.

Since

x = \log_2(a) = 3

, we have

a = 2^3 = 8

. Since

y = \log_2(b) = 2

, we have

b = 2^2 = 4

.

The definition of a logarithm

\log_b(z) = w

is equivalent to

z = b^w

.

5

Calculate the sum of

a

and

b

.

a + b = 8 + 4 = 12

.

To find the final value requested by the question stem.

Key Concept

Solving systems of logarithmic equations using logarithm properties and exponential conversions

Question 6Question →

If $\log_3(2x - 1) = 2$ , what is the value of $x$ ?

Show answer & explanation

Answer: 5

Answer

The value of

x

is

5

.

Converting the logarithmic equation

\log_3(2x - 1) = 2

to its equivalent exponential form yields

3^2 = 2x - 1

. Simplifying the exponent gives

9 = 2x - 1

. Adding

1

to both sides results in

10 = 2x

, and dividing by

2

gives

x = 5

.

Step-by-Step Solution

1

Rewrite the logarithmic equation in exponential form.

2x - 1 = 3^2

By definition,

\log_b(y) = z

is equivalent to

b^z = y

.

2

Evaluate the exponent

3^2

.

2x - 1 = 9

Calculating

3

squared yields

9

.

3

Solve the linear equation for

x

.

2x = 10

, which simplifies to

x = 5

Adding

1

to both sides and then dividing by

2

isolates

x

.

Key Concept

Converting logarithmic equations to exponential equations

Estimated Time:45s

Question 7Question →

If $\log_4 x = \frac{3}{2}$ , what is the value of $x$ ?

6

8

12

64

\frac{81}{16}

Show answer & explanation

Answer: $8$

Answer

The correct answer is

8

.

To solve the equation

\log_4 x = \frac{3}{2}

, we apply the definition of a logarithm to rewrite it in exponential form:

x = 4^{3/2}

. We then evaluate the exponent by first taking the square root of

4

, which is

2

, and then cubing it to get

2^3 = 8

. Thus, the value of

x

is

8

.

Step-by-Step Solution

1

Rewrite the logarithmic equation in its equivalent exponential form.

x = 4^{3/2}

By definition, a logarithmic equation of the form

\log_b a = c

is equivalent to the exponential equation

b^c = a

.

2

Evaluate the exponential expression

4^{3/2}

.

x = 8

The fractional exponent can be simplified by taking the square root of the base first, which is

\sqrt{4} = 2

, and then raising the result to the power of the numerator, giving

2^3 = 8

.

Key Concept

Converting logarithmic equations to exponential form and evaluating fractional exponents

Alternative Method

You can solve this by substituting the answer choices back into the equation. For example, testing the value

8

gives

\log_4 8 = \log_4 (2^3) = 3 \log_4 2

. Since

2

is the square root of

4

,

\log_4 2 = \frac{1}{2}

. Therefore,

3 \times \frac{1}{2} = \frac{3}{2}

, confirming that

8

is the correct solution.

Estimated Time:45s

Question 8Question →

If $x$ and $y$ are real numbers such that $2^x \cdot 3^y = 24$ and $3^x \cdot 2^y = 54$ , what is the value of $x^2 + y^2$ ?

Show answer & explanation

Answer: 10

Answer

10

Multiplying the two equations gives

(2^x \cdot 3^y)(3^x \cdot 2^y) = 24 \cdot 54 \implies 6^{x+y} = 1296 = 6^4

, which yields

x + y = 4

. Dividing the first equation by the second equation gives \frac{2^x \cdot 3^y}{3^x \cdot 2^y} = \frac{24}{54} \implies (\frac{2}{3})^{x-y} = \frac{4}{9} = (\frac{2}{3})^2

, which yields

x - y = 2

. Solving the system of equations

x + y = 4

and

x - y = 2

gives

x = 3

and

y = 1

. The value of

x^2 + y^2

is therefore

3^2 + 1^2 = 10$.

Step-by-Step Solution

1

Multiply the two equations together.

6^{x+y} = 1296

, which simplifies to

x + y = 4

.

Multiplying the equations groups bases of 2 and 3 together to form base 6, allowing us to find the sum of the variables.

2

Divide the first equation by the second equation.

(\frac{2}{3})^{x-y} = \frac{4}{9}

, which simplifies to

x - y = 2

.

Dividing the equations groups bases of 2 and 3 to form base

2/3

, allowing us to find the difference of the variables.

3

Solve the system of equations for

x

and

y

.

x = 3

and

y = 1

.

Solving the linear system of equations

x + y = 4

and

x - y = 2

gives the individual values of

x

and

y

.

4

Calculate

x^2 + y^2

.

10

Substitute the values of

x

and

y

into the target expression.

Key Concept

Solving systems of exponential equations using properties of exponents and bases

Alternative Method

Take the logarithm of both sides of each equation to convert them into a system of linear equations in terms of

x

and

y

:

x \log 2 + y \log 3 = \log 24

and

x \log 3 + y \log 2 = \log 54

. Solving this system using elimination or substitution yields

x = 3

and

y = 1

, so

x^2 + y^2 = 10

.

Estimated Time:2m 30s

Question 9Question →

If $x$ is a positive real number such that $\log_2(x) \cdot \log_3(x) - \log_2(x) - \log_3(x) = 1$ , what is the product of all possible real values of $x$ ?

\frac{1}{6}

5

6

\frac{1}{5}

12

Show answer & explanation

Answer: $6$

Answer

6

The correct answer is 6. By converting the equation using the change-of-base formula into natural logarithms, we obtain the quadratic equation

(\ln x)^2 - (\ln 2 + \ln 3)\ln x - \ln 2 \cdot \ln 3 = 0

. By setting

w = \ln x

, we have a quadratic in terms of

w

. The product of the two solutions

x_1

and

x_2

is

x_1 x_2 = e^{w_1} e^{w_2} = e^{w_1 + w_2}

. Using Vieta's formulas, the sum of the roots

w_1 + w_2 = \ln 2 + \ln 3 = \ln 6

. Therefore, the product of the solutions is

e^{\ln 6} = 6

.

Step-by-Step Solution

1

Rewrite the logarithmic equation using a common base.

Using the change-of-base formula

\log_b(a) = \frac{\ln a}{\ln b}

, the equation becomes:

\left(\frac{\ln x}{\ln 2}\right)\left(\frac{\ln x}{\ln 3}\right) - \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 3} = 1

This allows all logarithmic terms to be expressed in terms of the natural logarithm, making it easier to solve.

2

Eliminate denominators by multiplying the equation by

\ln 2 \cdot \ln 3

.

(\ln x)^2 - (\ln 3)\ln x - (\ln 2)\ln x = \ln 2 \cdot \ln 3

Multiplying through by the common denominator simplifies the rational equation into a standard polynomial form.

3

Rearrange the equation into a standard quadratic form.

(\ln x)^2 - (\ln 2 + \ln 3)\ln x - \ln 2 \cdot \ln 3 = 0

Grouping the coefficients of

\ln x

reveals a quadratic equation of the form

Aw^2 + Bw + C = 0

, where

w = \ln x

.

4

Define the relationship between the roots of the quadratic equation and the values of

x

.

Let the two real roots of the quadratic equation be

w_1

and

w_2

. These correspond to the solutions for

x

, namely

x_1 = e^{w_1}

and

x_2 = e^{w_2}

. The product of the solutions is

x_1 \cdot x_2 = e^{w_1} \cdot e^{w_2} = e^{w_1 + w_2}

.

To find the product of the solutions for

x

, we must compute the exponential of the sum of the roots of the quadratic equation.

5

Apply Vieta's formulas to find the sum of the roots

w_1 + w_2

.

The sum of the roots is

w_1 + w_2 = -\frac{B}{A} = -\frac{-(\ln 2 + \ln 3)}{1} = \ln 2 + \ln 3

. Using the logarithmic product rule, this simplifies to

\ln(2 \cdot 3) = \ln 6

.

Vieta's formulas state that the sum of the roots of

Aw^2 + Bw + C = 0

is

-\frac{B}{A}

.

6

Calculate the product of the real values of

x

.

x_1 \cdot x_2 = e^{\ln 6} = 6

Substituting the sum of the roots back into the exponent gives the final product.

Key Concept

Solving equations involving logarithmic properties, change of base, and relating quadratic roots to exponential functions

Question 10Question →

For all positive real numbers $x$ , which of the following expressions is equivalent to $\log_3(27x^4) - \log_3(3x^2)$ ?

2 + 6\log_3(x)

\log_3(24x^2)

2 + 2\log_3(x)

4\log_3(x)

3 + 2\log_3(x)

Show answer & explanation

Answer: $2 + 2\log_3(x)$

Answer

2 + 2\log_3(x)

The correct answer is found by applying the quotient property of logarithms to combine the terms, yielding

\log_3(9x^2)

. Then, applying the product property splits this into

\log_3(9) + \log_3(x^2)

. Finally, evaluating

\log_3(9) = 2

and using the power property to rewrite

\log_3(x^2)

as

2\log_3(x)

yields the simplified expression

2 + 2\log_3(x)

.

Step-by-Step Solution

1

Apply the quotient property of logarithms:

\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)

.

\log_3\left(\frac{27x^4}{3x^2}\right)

To combine the two logarithmic terms into a single logarithm.

2

Simplify the algebraic expression inside the logarithm.

\log_3(9x^2)

Dividing the coefficients (

27 \div 3 = 9

) and subtracting the exponents of the variable

x

(

4 - 2 = 2

).

3

Apply the product property of logarithms:

\log_b(CD) = \log_b(C) + \log_b(D)

.

\log_3(9) + \log_3(x^2)

To separate the constant and variable parts of the logarithmic argument.

4

Evaluate the numerical logarithm and apply the power property of logarithms:

\log_b(y^k) = k\log_b(y)

.

2 + 2\log_3(x)

Since

3^2 = 9

,

\log_3(9) = 2

, and the exponent of

x

can be moved in front of the logarithm as a multiplier.

Key Concept

Logarithmic and Exponential Expressions and Equations

Alternative Method

Alternatively, expand each logarithm first using the product and power properties of logarithms:
1.

\log_3(27x^4) = \log_3(27) + \log_3(x^4) = 3 + 4\log_3(x)

2.

\log_3(3x^2) = \log_3(3) + \log_3(x^2) = 1 + 2\log_3(x)

Subtracting the second expanded expression from the first yields:

(3 + 4\log_3(x)) - (1 + 2\log_3(x)) = 3 - 1 + 4\log_3(x) - 2\log_3(x) = 2 + 2\log_3(x)

.

Estimated Time:1m 30s

Question 11Question →

If $\log_3(x + 4) = 4$ , what is the value of $x$ ?

Show answer & explanation

Answer: 77

Answer

The value of

x

is 77.

The correct answer is 77. To solve the equation

\log_3(x + 4) = 4

, convert the equation from its logarithmic form to its exponential form. Since

\log_b(a) = c

means

b^c = a

, the equation becomes

3^4 = x + 4

. Calculating

3^4

yields 81. Thus,

81 = x + 4

. Subtracting 4 from both sides gives

x = 77

.

Step-by-Step Solution

1

Rewrite the logarithmic equation in exponential form.

x + 4 = 3^4

By definition,

\log_b(a) = c

is equivalent to

b^c = a

.

2

Calculate the value of the exponential expression.

3^4 = 81

3^4 = 3 \times 3 \times 3 \times 3 = 81

.

3

Solve for

x

by isolating the variable.

x = 77

Subtract 4 from both sides of the equation:

81 - 4 = 77

.

Key Concept

Converting logarithmic equations to exponential form

Question 12Question →

If $x$ and $y$ are positive real numbers such that $\log_x(y) = 2$ and $\log_4(x) + \log_2(y) = 5$ , what is the value of $y$ ?

Show answer & explanation

Answer: 16

Answer

16

Applying the definition of logarithms to

\log_x(y) = 2

yields

y = x^2

. Substituting this relationship into the second equation gives

\log_4(x) + \log_2(x^2) = 5

. Using the change of base formula, we rewrite

\log_4(x)

as

\frac{\log_2(x)}{\log_2(4)} = \frac{1}{2}\log_2(x)

, and using the power property, we rewrite

\log_2(x^2)

as

2\log_2(x)

. Combining the terms results in

\frac{5}{2}\log_2(x) = 5

, which simplifies to

\log_2(x) = 2

. Converting this back to exponential form gives

x = 2^2 = 4

. Substituting this value back into the relation

y = x^2

yields

y = 4^2 = 16

.

Step-by-Step Solution

1

Apply the definition of a logarithm to the first equation.

y = x^2

By definition,

\log_b(a) = c

is equivalent to

b^c = a

.

2

Substitute

y = x^2

into the second equation.

\log_4(x) + \log_2(x^2) = 5

Substitution reduces the system to a single equation in terms of

x

.

3

Express both logarithmic terms using base 2 properties.

\frac{1}{2}\log_2(x) + 2\log_2(x) = 5

The change of base formula gives

\log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{1}{2}\log_2(x)

, and the power property gives

\log_2(x^2) = 2\log_2(x)

.

4

Combine the coefficients and solve for

x

.

\log_2(x) = 2 \Rightarrow x = 2^2 = 4

Adding the coefficients gives

\frac{5}{2}\log_2(x) = 5

. Multiplying by

\frac{2}{5}

isolates

\log_2(x)

to solve for

x

.

5

Calculate the value of

y

using the relation from Step 1.

y = 4^2 = 16

Since

y = x^2

and

x = 4

, evaluating the square gives the final answer.

Key Concept

Solving systems of logarithmic equations using base conversion and definition of logarithms

Estimated Time:2m 0s

Question 13Question →

If $a$ , $b$ , and $c$ are positive real numbers greater than $1$ such that $\log_b(a) = \frac{3}{2}$ and $\log_c(b) = \frac{4}{3}$ , what is the value of $\log_a(abc)$ ?

\frac{1}{3}

\frac{13}{6}

\frac{23}{9}

3

\frac{9}{2}

Show answer & explanation

Answer: $\frac{13}{6}$

Answer

\frac{13}{6}

The correct answer is obtained by expanding

\log_a(abc)

into

\log_a(a) + \log_a(b) + \log_a(c)

. Since

\log_a(a) = 1

and

\log_a(b)

is the reciprocal of

\log_b(a)

, we find

\log_a(b) = \frac{2}{3}

. Using the chain rule of change of base, we calculate

\log_a(c) = \log_a(b) \cdot \log_b(c) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}

. Adding these three parts together yields

1 + \frac{2}{3} + \frac{1}{2} = \frac{13}{6}

.

Step-by-Step Solution

1

Apply the logarithmic product rule to expand the target expression.

\log_a(abc) = \log_a(a) + \log_a(b) + \log_a(c)

The logarithm of a product is equal to the sum of the logarithms of its individual factors.

2

Evaluate the first two terms of the expanded expression.

\log_a(a) = 1

, and since

\log_b(a) = \frac{3}{2}

, we have

\log_a(b) = \frac{1}{\log_b(a)} = \frac{2}{3}

.

The logarithm of a base to itself is always

1

, and the base reciprocal property states that

\log_x(y) = \frac{1}{\log_y(x)}

.

3

Evaluate the third term

\log_a(c)

using base properties and the change of base formula.

Since

\log_c(b) = \frac{4}{3}

, we have

\log_b(c) = \frac{3}{4}

. Then,

\log_a(c) = \log_a(b) \cdot \log_b(c) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}

.

By applying the change of base formula, we can rewrite

\log_a(c)

in terms of base

b

as

\frac{\log_b(c)}{\log_b(a)} = \log_a(b) \cdot \log_b(c)

.

4

Sum the three evaluated logarithmic terms together.

1 + \frac{2}{3} + \frac{1}{2} = \frac{6}{6} + \frac{4}{6} + \frac{3}{6} = \frac{13}{6}

Combining the values of the individual terms gives the final value of the expanded expression.

Key Concept

Properties of Logarithms and Change of Base Formula

Estimated Time:2m 0s

Question 14Question →

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

-\frac{3}{8}

\frac{1}{2}

\frac{9}{8}

1

-\frac{3}{4}

Show answer & explanation

Answer: $\frac{9}{8}$

Answer

The correct answer is

\frac{9}{8}

.

By writing both sides of the equation with a common base of

2

, we get

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

. Applying the power-to-a-power exponent rule, this simplifies to

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

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

6x-3 = -2x+6

. Adding

2x

and

3

to both sides results in

8x = 9

, which gives

x = \frac{9}{8}

.

Step-by-Step Solution

1

Express both bases as powers of

2

.

8 = 2^3

and

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

, so the equation becomes

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

.

Finding a common base allows us to equate the exponents directly.

2

Apply the power-to-a-power rule

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

to simplify the exponents.

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

.

This simplifies the exponential expressions on both sides of the equation.

3

Equate the exponents and solve for

x

.

6x - 3 = -2x + 6 \Rightarrow 8x = 9 \Rightarrow x = \frac{9}{8}

.

Since the bases are equal, their exponents must be equal.

Key Concept

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

Estimated Time:1m 30s

Question 15Question →

What is the product of all real values of $x$ that satisfy the equation $\log_3(x) - 6\log_x(3) = 1$ ?

Show answer & explanation

Answer: 3

Answer

The product of all real values of

x

that satisfy the equation is 3.

By applying the change-of-base formula, the equation becomes

\log_3(x) - \frac{6}{\log_3(x)} = 1

. Substituting

y = \log_3(x)

leads to

y^2 - y - 6 = 0

, which has roots

y = 3

and

y = -2

. These roots correspond to

x = 3^3 = 27

and

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

. Both solutions are valid because they are positive and do not equal 1. The product of these solutions is

27 \times \frac{1}{9} = 3

. Alternatively, using Vieta's formulas, the sum of the roots of the quadratic equation is

y_1 + y_2 = 1

. The product of the solutions is

x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3

.

Step-by-Step Solution

1

Apply the change-of-base formula to rewrite the variable base term.

\log_3(x) - \frac{6}{\log_3(x)} = 1

This expresses the equation in terms of logarithms with the same base.

2

Use substitution to convert the equation into a quadratic form.

y - \frac{6}{y} = 1 \Rightarrow y^2 - y - 6 = 0

where

y = \log_3(x)

Substitution simplifies the logarithmic equation into a polynomial equation.

3

Solve the quadratic equation by factoring.

(y-3)(y+2) = 0 \Rightarrow y = 3

or

y = -2

Finding the roots for

y

is the intermediate step to solving for

x

.

4

Back-substitute to find the values of

x

.

x = 3^3 = 27

and

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

Converting from logarithmic form back to exponential form yields the values of

x

.

5

Multiply the solutions together.

27 \times \frac{1}{9} = 3

The question asks for the product of all real solutions.

Key Concept

Solving logarithmic equations using the change-of-base formula and quadratic substitution.

Alternative Method

Instead of solving for individual values of

x

, note that if

y_1

and

y_2

are the roots of the quadratic equation

y^2 - y - 6 = 0

, then

y_1 + y_2 = 1

by Vieta's formulas. Since

x_1 = 3^{y_1}

and

x_2 = 3^{y_2}

, the product of the solutions is

x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3

.

Estimated Time:2m 0s

Question 16Question →

For all positive real numbers $x$ and $y$ , which of the following expressions is equivalent to $\log_3(9x^4 y^{-2})$ ?

9 + 4\log_3(x) - 2\log_3(y)

2 + 4\log_3(x) + 2\log_3(y)

2 + 4\log_3(x) - 2\log_3(y)

\frac{2 + 4\log_3(x)}{2\log_3(y)}

2 + \log_3(4x) - \log_3(2y)

Show answer & explanation

Answer: $2 + 4\log_3(x) - 2\log_3(y)$

Answer

2 + 4\log_3(x) - 2\log_3(y)

The correct answer is correct because applying the product and power properties of logarithms allows us to expand the argument

9x^4 y^{-2}

. Specifically, the expression can be written as

\log_3(9) + \log_3(x^4) + \log_3(y^{-2})

. Since

3^2 = 9

, the term

\log_3(9)

evaluates to

2

. Applying the power rule

\log_b(M^k) = k\log_b(M)

to the remaining terms yields

4\log_3(x)

and

-2\log_3(y)

. Combining these terms results in the equivalent expression.

Step-by-Step Solution

1

Use the product rule of logarithms,

\log_b(MN) = \log_b(M) + \log_b(N)

, to separate the factors in the argument.

\log_3(9x^4 y^{-2}) = \log_3(9) + \log_3(x^4) + \log_3(y^{-2})

This allows the expression to be broken down into individual terms for each base and variable.

2

Evaluate the constant logarithm

\log_3(9)

. Since

3^2 = 9

, this simplifies to

2

.

\log_3(9) = 2

To simplify the numerical term to its integer value.

3

Use the power rule of logarithms,

\log_b(M^k) = k\log_b(M)

, to bring the exponents of the variables to the front of each logarithm.

\log_3(x^4) = 4\log_3(x)

and

\log_3(y^{-2}) = -2\log_3(y)

To isolate the variable terms inside simpler logarithmic expressions.

4

Combine all the simplified parts into a single expression.

2 + 4\log_3(x) - 2\log_3(y)

To write the final expanded equivalent expression.

Key Concept

Logarithmic properties, including product, power, and evaluation of basic log terms.

Alternative Method

Substitute test values for the variables. For example, let

x = 3

and

y = 3

. The original expression evaluates to

\log_3(9 \cdot 3^4 \cdot 3^{-2}) = \log_3(81) = 4

. Substituting

x = 3

and

y = 3

into the correct expression yields

2 + 4\log_3(3) - 2\log_3(3) = 2 + 4(1) - 2(1) = 4

, which matches the original expression's value.

Estimated Time:1m 0s

Question 17Question →

A certain radioactive isotope decays according to the formula $N(t) = N_0 \cdot 2^{-t/8}$ , where $N_0$ is the initial amount of the isotope and $t$ is the time in years. If a sample initially contains $120$ grams of the isotope, how many years will it take for the amount of the isotope to decay to $15$ grams?

Show answer & explanation

Answer: 24

Answer

24

The correct answer is 24 because substituting the initial value of 120 and the final value of 15 into the equation yields

15 = 120 \cdot 2^{-t/8}

. Dividing both sides by 120 gives

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

, which can be rewritten as

2^{-3} = 2^{-t/8}

. Setting the exponents equal to each other gives

-3 = -t/8

, and solving for

t

yields 24.

Step-by-Step Solution

1

Substitute the given values into the decay formula.

15 = 120 \cdot 2^{-t/8}

The initial amount

N_0

is

120

grams and the final amount

N(t)

is

15

grams.

2

Isolate the exponential term.

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

Divide both sides by

120

. Since

\frac{15}{120}

reduces to

\frac{1}{8}

, this isolates the base

2

term.

3

Write the fraction as a power with base 2.

2^{-3} = 2^{-t/8}

Using exponent rules,

\frac{1}{8} = \frac{1}{2^3} = 2^{-3}

.

4

Equate the exponents and solve for

t

.

t = 24

Since the bases are equal, the exponents must be equal, so

-3 = -\frac{t}{8}

which gives

t = 24

.

Key Concept

Solving exponential equations using a common base.

Question 18Question →

If $x$ is a real number greater than 2 such that $\log_2(x^2 - 4) - \log_2(x - 2) = 3$ , what is the value of $x$ ?

1

4

6

7

16

Show answer & explanation

Answer: 6

Answer

The value of

x

is 6.

Applying the quotient property of logarithms simplifies the equation to

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

. Factoring the numerator as

(x-2)(x+2)

and canceling the common factor

(x-2)

yields

\log_2(x+2) = 3

. Rewriting this in exponential form gives

x + 2 = 2^3 = 8

, which yields

x = 6

.

Step-by-Step Solution

1

Combine the logarithmic terms on the left side of the equation using the quotient property of logarithms:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

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

Consolidating the two logarithmic terms simplifies the equation into a single logarithmic term.

2

Factor the numerator as a difference of squares and simplify the rational expression.

\log_2(x + 2) = 3

(since

x > 2

,

x - 2 \neq 0

)

Factoring

x^2 - 4

as

(x - 2)(x + 2)

allows the cancellation of the common factor

(x - 2)

.

3

Convert the logarithmic equation into its equivalent exponential form:

\log_b Y = c \Rightarrow Y = b^c

.

x + 2 = 2^3

This removes the logarithm and sets up a linear equation to solve.

4

Evaluate the exponent and solve for

x

.

x + 2 = 8 \Rightarrow x = 6

Evaluating

2^3

as 8 and subtracting 2 from both sides isolates

x

.

Key Concept

Solving logarithmic equations using properties of logarithms and algebraic factoring

Estimated Time:1m 30s

Question 19Question →

If $a > 1.5$ is a real number that satisfies the equation $4^a - 3(2^{a+1}) + 8 = 0$ , what is the value of $a$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

Substituting

u = 2^a

turns the equation into

u^2 - 6u + 8 = 0

. Factoring gives

(u-2)(u-4)=0

, meaning

u=2

or

u=4

. Reversing the substitution gives

2^a = 2 \implies a=1

and

2^a = 4 \implies a=2

. Since

a > 1.5

, the correct value is 2.

Step-by-Step Solution

1

Express the equation in terms of base 2.

(2^a)^2 - 6(2^a) + 8 = 0

Since 4 is

2^2

and

2^{a+1} = 2 \cdot 2^a

, expressing all terms in base 2 allows for algebraic substitution.

2

Substitute

u = 2^a

to form a quadratic equation.

u^2 - 6u + 8 = 0

Substitution simplifies the exponential equation into a standard quadratic form.

3

Factor the quadratic equation.

(u-2)(u-4) = 0

Factoring allows us to find the roots of the quadratic equation.

4

Solve for the variable

a

.

a = 1

or

a = 2

Solving

2^a = 2

yields

a = 1

, and solving

2^a = 4

yields

a = 2

.

5

Apply the given constraint on

a

.

a = 2

The problem states that

a > 1.5

, so

a = 1

is discarded and

a = 2

is the correct value.

Key Concept

Solving exponential equations using quadratic substitution

Alternative Method

Instead of using substitution, test values for

a

. Since

a > 1.5

, testing small integer values starting with

a=2

shows

4^2 - 3(2^3) + 8 = 16 - 24 + 8 = 0

, validating that

a=2

is the solution.

Estimated Time:1m 30s

Question 20Question →

For all real numbers $x$ and $y$ such that $x > y > 0$ , which of the following expressions is equivalent to $\log_5(x^2 - y^2) - \log_5(x - y)$ ?

\log_5(x-y)

\log_5(x^2 - y^2 - x + y)

\log_5(x) - \log_5(y)

\log_5(x+y)

x + y

Show answer & explanation

Answer: $\log_5(x+y)$

Answer

The expression is equivalent to \log_5(x + y)

The correct answer is found by using the quotient property of logarithms to combine the terms:

\log_5(x^2 - y^2) - \log_5(x - y) = \log_5\left(\frac{x^2 - y^2}{x - y}\right)

. Factoring the difference of squares in the numerator yields

\log_5\left(\frac{(x-y)(x+y)}{x-y}\right)

. Canceling the common factor

(x-y)

simplifies the expression to

\log_5(x+y)

.

Step-by-Step Solution

1

Combine the logarithms using the quotient property of logarithms:

\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)

.

\log_5\left(\frac{x^2 - y^2}{x - y}\right)

This simplifies the difference between two logarithms with the same base into a single logarithm.

2

Factor the difference of squares in the numerator:

x^2 - y^2 = (x - y)(x + y)

.

\log_5\left(\frac{(x - y)(x + y)}{x - y}\right)

This allows for the cancellation of common factors in the fraction.

3

Cancel the common factor

(x - y)

from both the numerator and the denominator.

\log_5(x + y)

Since

x > y > 0

, the term

x - y

is non-zero, making the division valid.

Key Concept

Quotient property of logarithms and difference of squares factoring

Alternative Method

Instead of applying the quotient property first, one can factor the argument of the first term using the difference of squares:

\log_5(x^2 - y^2) = \log_5((x - y)(x + y))

. Next, apply the product property of logarithms to split this term:

\log_5(x - y) + \log_5(x + y)

. Substituting this back into the original expression gives

\log_5(x - y) + \log_5(x + y) - \log_5(x - y)

, which simplifies directly to

\log_5(x + y)

.

Estimated Time:1m 0s

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