Algebraic Exponents and Radicals Practice Questions — GRE General Test

Question 1Question →

If $b > 1$ , which of the following expressions are equivalent to $\frac{\sqrt{b^3 \cdot \sqrt[3]{b^2}}}{b^{-\frac{1}{6}}}$ ? Select all that apply.

Select all that apply

(\sqrt[3]{b})^6

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}

\sqrt{b^4 + b^2}

b^{\frac{11}{6}} \cdot b^{-\frac{1}{6}}

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

Show answer & explanation

Answer: $(\sqrt[3]{b})^6$ ; $\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}$ ; $\left(b^{-\frac{1}{2}}\right)^{-4}$

Answer

The equivalent expressions are

(\sqrt[3]{b})^6

,

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}

, and

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

.

Simplifying the original expression yields

\frac{\sqrt{b^3 \cdot b^{2/3}}}{b^{-1/6}} = \frac{b^{11/6}}{b^{-1/6}} = b^{11/6 - (-1/6)} = b^{12/6} = b^2

. The expression

(\sqrt[3]{b})^6

equals

b^{6/3} = b^2

. The expression

\frac{b^{7/3}}{b^{1/3}}

equals

b^{7/3 - 1/3} = b^2

. The expression

\left(b^{-1/2}\right)^{-4}

equals

b^{(-1/2)(-4)} = b^2

. All three equal

b^2

.

Step-by-Step Solution

1

Simplify the expression inside the square root in the numerator.

b^3 \cdot \sqrt[3]{b^2} = b^3 \cdot b^{\frac{2}{3}} = b^{3 + \frac{2}{3}} = b^{\frac{11}{3}}

Convert radical to rational exponent and use product rule for exponents.

2

Apply the square root to the numerator.

\sqrt{b^{\frac{11}{3}}} = \left(b^{\frac{11}{3}}\right)^{\frac{1}{2}} = b^{\frac{11}{6}}

Taking the square root is equivalent to raising to the power of

\frac{1}{2}

.

3

Divide by the denominator.

\frac{b^{\frac{11}{6}}}{b^{-\frac{1}{6}}} = b^{\frac{11}{6} - \left(-\frac{1}{6}\right)} = b^{\frac{12}{6}} = b^2

Apply quotient rule for exponents: subtract the exponent of the denominator from the numerator.

4

Evaluate each given option to determine which simplify to

b^2

.

The expressions

(\sqrt[3]{b})^6 = b^2

,

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}} = b^2

, and

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

are all equivalent.

Matching each simplified candidate expression to the target simplified value

b^2

.

Key Concept

Simplification of nested algebraic radicals and rational exponents using exponent rules

Estimated Time:2m 0s

Question 2Question →

If $x$ and $y$ are positive integers such that $2^{x+1} + 2^x = 3^{y+2} - 3^y$ , what is the value of $x + y$ ?

Show answer & explanation

Answer: 4

Answer

The value of

x + y

is

4

.

Factoring the left side gives

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

, while factoring the right side gives

3^y(9 - 1) = 8 \cdot 3^y = 2^3 \cdot 3^y

. Equating the two expressions gives

3 \cdot 2^x = 2^3 \cdot 3^y

. Rearranging terms to separate bases yields

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

. Because 2 and 3 share no common prime factors, this equality holds for integers if and only if both exponents are equal to 0. Solving

x - 3 = 0

gives

x = 3

, and solving

y - 1 = 0

gives

y = 1

. Both are positive integers. Thus,

x + y = 3 + 1 = 4

.

Step-by-Step Solution

1

Factor out common terms on both sides of the equation.

2^x(2 + 1) = 3^y(9 - 1) \implies 3 \cdot 2^x = 8 \cdot 3^y

Factoring simplifies sums of powers with identical bases.

2

Rewrite integers using prime factorizations and re-group bases.

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

Dividing both sides by

2^3 \cdot 3

separates the base-2 and base-3 exponential terms.

3

Set each exponent to zero using prime independence.

x - 3 = 0 \implies x = 3

and

y - 1 = 0 \implies y = 1

Powers of distinct prime numbers 2 and 3 can only be equal if both powers equal

1

(

2^0 = 3^0 = 1

).

4

Calculate the required sum

x + y

.

3 + 1 = 4

Evaluates the requested combined value of the variables.

Key Concept

Solving exponential equations involving distinct prime bases through factoring and exponent properties.

Question 3Question →

If

x

is a real number satisfying the equation

\frac{\left(2^{3x+2} + 2^{3x}\right)^3}{\left(4^{2x+1} - 4^{2x}\right)^2} = \frac{2000}{9}

what is the value of

x

?

Show answer & explanation

Answer: 4

Answer

4

Factoring out

2^{3x}

in the numerator gives

2^{3x}(2^2 + 1) = 5 \cdot 2^{3x}

. Cubing this yields

125 \cdot 2^{9x}

. In the denominator, rewriting

4^{2x}

as

2^{4x}

and factoring gives

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

. Squaring this yields

9 \cdot 2^{8x}

. Taking the quotient gives

\frac{125}{9} \cdot 2^x

. Setting this equal to

\frac{2000}{9}

leads directly to

125 \cdot 2^x = 2000 \implies 2^x = 16 \implies x = 4

.

Step-by-Step Solution

1

Factor out common exponential terms inside the parentheses

Numerator inside becomes

5 \cdot 2^{3x}

and denominator inside becomes

3 \cdot 2^{4x}

Factoring out

2^{3x}

from

2^{3x+2} + 2^{3x}

isolates the constant multiplier

(4+1)

, and expressing

4^{2x}

as

2^{4x}

allows base unification.

2

Raise the simplified terms to their respective outer powers

Numerator becomes

125 \cdot 2^{9x}

and denominator becomes

9 \cdot 2^{8x}

Using power rules

(a \cdot b)^n = a^n b^n

and

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

.

3

Simplify the fraction by subtracting exponents of like bases

The left side simplifies to

\frac{125}{9} \cdot 2^x

\ me{2^{9x}}{2^{8x}} = 2^{9x-8x} = 2^x

using the quotient rule for exponents.

4

Solve the resulting single-variable exponential equation

2^x = 16

, which yields

x = 4

Multiplying both sides by

9

yields

125 \cdot 2^x = 2000

, so

2^x = 16 = 2^4

.

Key Concept

Factoring exponential expressions and applying power of a power and quotient rules

Question 4Question →

If $x$ is a positive real number, which of the following is equivalent to the expression $\frac{(x^3)^2 \cdot x^{-4}}{\sqrt{x^8}}$ ?

x^{-2}

x^2

x^{-3}

x^{-4}

x^0

Show answer & explanation

Answer: $x^{-2}$

Answer

The simplified expression is equivalent to

x^{-2}

.

Applying the rules of exponents systematically yields

(x^3)^2 = x^6

in the numerator, which combines with

x^{-4}

to give

x^2

. The denominator

\sqrt{x^8}

simplifies to

x^{8/2} = x^4

. Dividing

x^2

by

x^4

gives

x^{2-4} = x^{-2}

.

Step-by-Step Solution

1

Simplify the power raised to a power in the numerator.

(x^3)^2 = x^{3 \cdot 2} = x^6

When raising a power to a power, multiply the exponents:

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

.

2

Multiply the terms in the numerator.

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

When multiplying exponential expressions with the same base, add the exponents:

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

.

3

Simplify the radical in the denominator.

\sqrt{x^8} = (x^8)^{\frac{1}{2}} = x^{\frac{8}{2}} = x^4

Taking the square root of a non-negative term is equivalent to raising it to the power of

\frac{1}{2}

.

4

Divide the numerator by the denominator.

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

When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent: \frac{a^m}{a^n} = a^{m-n}.

Key Concept

Laws of Exponents and Radical Simplification

Estimated Time:45s

Question 5Question →

If $x > 0$ , which of the following is equivalent to the expression $\sqrt{9x^4 + 16x^4}$ ?

5x^2

7x^2

5x^4

7x^4

25x^2

Show answer & explanation

Answer: $5x^2$

Answer

The correct answer is

5x^2

.

First combine the terms under the square root to get

\sqrt{25x^4}

. Taking the square root yields

\sqrt{25} \cdot \sqrt{x^4} = 5x^2

. Thus,

5x^2

is the correct simplified expression.

Step-by-Step Solution

1

Combine like terms inside the square root.

9x^4 + 16x^4 = 25x^4

Since both terms have the same variable factor

x^4

, their coefficients can be added directly.

2

Apply the product rule for radicals:

\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}

.

\sqrt{25x^4} = \sqrt{25} \cdot \sqrt{x^4}

The square root of a product is equal to the product of the square roots.

3

Evaluate the square root of the coefficient and the variable term.

5 \cdot x^2 = 5x^2

\sqrt{25} = 5

and

\sqrt{x^4} = (x^4)^{1/2} = x^2

(since

x > 0$).

Key Concept

Simplifying algebraic radicals by combining like terms under the square root and applying power rules for exponents.

Question 6Question →

If $x > 1$ , which of the following expressions are equivalent to $\frac{\sqrt{x^5 \cdot x^3}}{(x^{-2})^{-1}}$ ? Select all such expressions.

Select all that apply

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

\sqrt{x^4 + x^4}

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

(x^{-1})^{-2} \cdot x^{-4}

\sqrt[4]{x^4}

Show answer & explanation

Answer: $(x^{\frac{2}{3}})^3$ ; $\frac{x^{-1}}{x^{-3}}$

Answer

The expressions equivalent to the target expression are

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

and

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

.

The original expression simplifies step by step to

x^2

. The expression

(x^{2/3})^3

equals

x^{(2/3)\cdot 3} = x^2

, and the expression

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

equals

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

. Both are equal to

x^2

.

Step-by-Step Solution

1

Simplify the numerator of the given expression

\sqrt{x^5 \cdot x^3} = \sqrt{x^{5+3}} = \sqrt{x^8} = x^{8/2} = x^4

Combine bases using product rule for exponents, then convert radical to fractional exponent.

2

Simplify the denominator of the given expression

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

Multiply exponents when raising a power to a power.

3

Divide the simplified numerator by the simplified denominator

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

Subtract denominator exponent from numerator exponent using the quotient rule.

4

Evaluate each choice against

x^2

Only

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

and

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

are equal to

x^2

.

Apply basic exponent rules to each proposed expression.

Key Concept

Algebraic Exponents and Radicals Rules

Question 7Question →

Which of the following values of $x$ are solutions to the equation $(x - 2)^{\frac{4}{3}} - 5(x - 2)^{\frac{2}{3}} + 4 = 0$ ? Select all that apply.

Select all that apply

-6

1

10

-2

6

Show answer & explanation

Answer: -6; 1; 10

Answer

The correct values of

x

that satisfy the equation are

-6

,

1

, and

10

.

Substituting

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

yields

u^2 - 5u + 4 = 0

, which factors as

(u - 1)(u - 4) = 0

, giving

u = 1

and

u = 4

. Solving

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

gives

(x - 2)^2 = 1 \implies x - 2 = \pm 1

, yielding

x = 3

and

x = 1

. Solving

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

gives

(x - 2)^2 = 64 \implies x - 2 = \pm 8

, yielding

x = 10

and

x = -6

. Thus, the values

-6

,

1

, and

10

are all valid solutions.

Step-by-Step Solution

1

Perform a substitution to rewrite the equation in quadratic form.

Let

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

. Then

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

, giving

u^2 - 5u + 4 = 0

.

Recognizing quadratic structure simplifies equations with rational exponents.

2

Solve the quadratic equation for

u

.

(u - 1)(u - 4) = 0 \implies u = 1

or

u = 4

.

Factoring determines the values of the substituted variable

u

.

3

Solve for

x

when

u = 1

.

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

or

x = 1

.

Raising both sides to the power of

\frac{3}{2}

requires taking both positive and negative roots because the numerator of the power is even.

4

Solve for

x

when

u = 4

.

(x - 2)^{\frac{2}{3}} = 4 \implies (x - 2)^2 = 4^3 = 64 \implies x - 2 = \pm 8 \implies x = 10

or

x = -6

.

Squaring and taking square roots yields two solutions,

10

and

-6

.

5

Match calculated solutions with the given choices.

The solutions present among the options are

-6

,

1

, and

10

.

Comparing all valid algebraic solutions to the available choices identifies all correct options.

Key Concept

Solving quadratic-form equations with fractional exponents and accounting for negative base branches when taking even roots.

Question 8Question →

If $a > 0$ , which of the following is equivalent to the expression $\sqrt{\frac{a^8 \cdot a^4}{a^{-2}}}$ ?

a^7

a^5

a^6

a^{14}

a^{10}

Show answer & explanation

Answer: $a^7$

Answer

a^7

Multiplying the terms in the numerator gives

a^{12}

. Dividing by

a^{-2}

gives

a^{12 - (-2)} = a^{14}

. Taking the square root of

a^{14}

gives

(a^{14})^{1/2} = a^7

.

Step-by-Step Solution

1

Simplify the numerator inside the square root using the product rule

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

.

a^8 \cdot a^4 = a^{8+4} = a^{12}

Powers with the same base are multiplied by adding their exponents.

2

Divide by the denominator using the quotient rule

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

.

\frac{a^{12}}{a^{-2}} = a^{12 - (-2)} = a^{14}

Dividing powers with the same base requires subtracting the lower exponent from the upper exponent.

3

Apply the fractional exponent rule for radicals

\sqrt{x} = x^{\frac{1}{2}}

.

\sqrt{a^{14}} = (a^{14})^{\frac{1}{2}} = a^{14 \cdot \frac{1}{2}} = a^7

Taking the square root of a power is equivalent to multiplying the exponent by

\frac{1}{2}

.

Key Concept

Simplifying expressions using exponent laws and fractional radical powers

Question 9Question →

If $2^{x + 3} \cdot 4^{x - 1} = 16^x$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1

Answer

The value of

x

is 1.

Rewriting

4^{x-1}

as

2^{2x-2}

and

16^x

as

2^{4x}

transforms the left side into

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

. Setting exponents equal gives

3x + 1 = 4x

, which simplifies to

x = 1

.

Step-by-Step Solution

1

Express all terms with a common base of 2

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

Converting

4

to

2^2

and

16

to

2^4

allows all terms to share the base 2.

2

Apply the power of a power rule

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

and the product rule

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

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

Multiplying exponents gives

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

and

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

. Adding exponents on the left gives

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

.

3

Equate the exponents and solve for

x

x = 1

Since

2^A = 2^B

implies

A = B

, setting

3x + 1 = 4x

directly yields

x = 1

.

Key Concept

Solving exponential equations using common bases and exponent properties

Question 10Question →

For all $x > 1$ , which of the following expressions is equivalent to $\frac{\sqrt{x^3 \cdot \sqrt[3]{x^2}}}{(x^{-1/3})^2}$ ?

x^{5/2}

x^{7/6}

x^{5/3}

x^{13/3}

x^{35/12}

Show answer & explanation

Answer: $x^{5/2}$

Answer

x^{5/2}

Converting all radical expressions into rational exponent form simplifies the numerator to

(x^{11/3})^{1/2} = x^{11/6}

and the denominator to

x^{-2/3}

. Applying the quotient rule

x^{11/6} / x^{-2/3} = x^{11/6 - (-2/3)}

yields

x^{15/6} = x^{5/2}

.

Step-by-Step Solution

1

Convert radicals in the numerator to fractional exponents and combine terms inside the square root

x^3 \cdot \sqrt[3]{x^2} = x^3 \cdot x^{2/3} = x^{3 + 2/3} = x^{11/3}

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

2

Apply the outer square root to the simplified expression in the numerator

\sqrt{x^{11/3}} = (x^{11/3})^{1/2} = x^{11/6}

Taking the square root is equivalent to raising an expression to the power of

1/2

.

3

Simplify the denominator using the power of a power rule

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

When raising a power to another power, multiply the exponents.

4

Divide the simplified numerator by the simplified denominator

\frac{x^{11/6}}{x^{-2/3}} = x^{11/6 - (-2/3)} = x^{11/6 + 4/6} = x^{15/6} = x^{5/2}

When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent.

Key Concept

Simplifying nested algebraic radicals using rational exponent laws

Estimated Time:1m 30s

Question 11Question →

For all positive real numbers $x$ and $y$ , which of the following expressions are equivalent to $\left(\frac{x^{-2} y^{3/2}}{\sqrt{x^4 y^{-1}}}\right)^{-1/2}$ ? Select all that apply.

Select all that apply

x^2 y^{-1}

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

\frac{y}{x^2}

\frac{x^3 y}{\sqrt{x^2 y^4}}

\frac{x^2}{\sqrt{y}}

Show answer & explanation

Answer: $x^2 y^{-1}$ ; $\sqrt{\frac{x^4}{y^2}}$ ; $\frac{x^3 y}{\sqrt{x^2 y^4}}$

Answer

The expressions equivalent to the given expression are

x^2 y^{-1}

,

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

, and

\frac{x^3 y}{\sqrt{x^2 y^4}}

.

Simplifying the original expression step-by-step yields

\frac{x^2}{y}

, which is equal to

x^2 y^{-1}

. Taking the square root of

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

gives

\frac{x^2}{y}

, and simplifying

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

also gives

\frac{x^2}{y}

. Thus, these three choices are equivalent to the target expression.

Step-by-Step Solution

1

Simplify the denominator inside the parentheses

\sqrt{x^4 y^{-1}} = (x^4 y^{-1})^{1/2} = x^2 y^{-1/2}

Apply the power of a product rule and principal square root properties for positive variables.

2

Simplify the fraction inside the parentheses

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

Subtract exponents of like bases when dividing.

3

Apply the outer exponent

-1/2

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

Multiply exponents when raising a power to a power.

4

Evaluate each option against

\frac{x^2}{y}

x^2 y^{-1}

,

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

, and

\frac{x^3 y}{\sqrt{x^2 y^4}}

all simplify to

\frac{x^2}{y}

.

Check algebraic equivalence for each provided choice.

Key Concept

Algebraic Exponents and Radicals

Question 12Question →

If

x

and

y

are positive real numbers satisfying the system of exponential radical equations

\sqrt{x\sqrt{y}} = 3

\sqrt{y\sqrt{x}} = 9

which of the following statements must be true? Select all that apply.

Select all that apply

xy = 81

x^{1/2} + y^{1/2} = 10

\sqrt{xy} = 27

x

has two valid real solutions,

x = 1

and

x = -1

y^x - x^y = 80

Show answer & explanation

Answer: $xy = 81$ ; $x^{1/2} + y^{1/2} = 10$ ; $y^x - x^y = 80$

Answer

The statements asserting that

xy = 81

,

x^{1/2} + y^{1/2} = 10

, and

y^x - x^y = 80

are true.

Solving the system of radical equations gives the unique positive solution pair

x = 1

and

y = 81

. Substituting these values into the statements shows that

xy = 1 \cdot 81 = 81

,

x^{1/2} + y^{1/2} = \sqrt{1} + \sqrt{81} = 10

, and

y^x - x^y = 81^1 - 1^{81} = 80

are all mathematically valid.

Step-by-Step Solution

1

Square both sides of each equation to remove the outer radicals

x\sqrt{y} = 9 \implies x^2 y = 81

and

y\sqrt{x} = 81 \implies y^2 x = 6561

Squaring

\sqrt{A} = B

yields

A = B^2

, and rewriting radicals as rational powers gives

x y^{1/2} = 9

and

y x^{1/2} = 81

.

2

Express

y

in terms of

x

from the first equation and substitute into the second

y = \frac{81}{x^2} \implies \left(\frac{81}{x^2}\right)^2 x = 6561 \implies \frac{6561}{x^3} = 6561

Substituting

y

eliminates the variable

y

, leaving a single equation in terms of

x

.

3

Solve for

x

and

y

x^3 = 1 \implies x = 1

, which gives

y = \frac{81}{1^2} = 81

Since

x > 0

is a positive real number,

x = 1

is the unique real solution, yielding

y = 81

.

4

Evaluate each given statement using

x = 1

and

y = 81

xy = 81

(True),

x^{1/2} + y^{1/2} = 1 + 9 = 10

(True),

\sqrt{xy} = 9 \neq 27

(False),

x = -1

is invalid (False),

81^1 - 1^{81} = 80

(True)

Direct calculation confirms which individual statements hold.

Key Concept

Simplifying nested radical equations using fractional exponent rules

Estimated Time:2m 0s

Question 13Question →

If $x$ is a real number such that $2^{x+1} + 2^{x-1} = 40$ , what is the value of the expression $(x+1)^{x-1}$ ?

65

64

125

243

15

Show answer & explanation

Answer: 125

Answer

125

Factoring out

2^x

from the given equation yields

2^x(2 + 0.5) = 40

, which gives

2.5 \cdot 2^x = 40

and

2^x = 16

. This determines that

x = 4

. Substituting

x = 4

into

(x+1)^{x-1}

gives

(4+1)^{3} = 5^3 = 125

.

Step-by-Step Solution

1

Rewrite the given exponential terms with a common power of 2.

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

and

2^{x-1} = 2^x \cdot 2^{-1} = \frac{2^x}{2}

.

Applying product rule for exponents allows factoring out

2^x

.

2

Factor out

2^x

and solve for

x

.

2^x \left(2 + \frac{1}{2}\right) = 40 \implies 2^x \cdot \frac{5}{2} = 40 \implies 2^x = 16 \implies x = 4

.

Combining fractional coefficients isolates the exponential term

2^x

.

3

Substitute

x = 4

into the target expression

(x+1)^{x-1}

.

(4+1)^{4-1} = 5^3 = 125

.

Simplifying the base and exponent yields the final numerical value.

Key Concept

Solving exponential equations using power distribution rules and factoring.

Question 14Question →

What value of $x$ satisfies the exponential equation $25^x - 4 \cdot 5^{x+1} = 125$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is

2

.

Rewriting

25^x

as

(5^x)^2

and

4 \cdot 5^{x+1}

as

20 \cdot 5^x

transforms the equation into

(5^x)^2 - 20(5^x) - 125 = 0

. Substituting

u = 5^x

produces

u^2 - 20u - 125 = 0

, which factors into

(u - 25)(u + 5) = 0

. Because

5^x

must be greater than zero for all real values of

x

,

u = -5

yields no valid real solution. Thus,

5^x = 25 = 5^2

, giving

x = 2

.

Step-by-Step Solution

1

Convert exponential expressions to a common base of 5.

25^x = (5^2)^x = (5^x)^2

and

4 \cdot 5^{x+1} = 4 \cdot 5 \cdot 5^x = 20 \cdot 5^x

.

Applying exponent laws

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

and

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

expresses terms in quadratic form with respect to

5^x

.

2

Formulate and factor the quadratic equation in terms of

u = 5^x

.

u^2 - 20u - 125 = 0 \implies (u - 25)(u + 5) = 0

, yielding

u = 25

or

u = -5

.

The equation reduces to standard quadratic form, which factors easily.

3

Solve for

x

while rejecting non-viable real roots.

5^x = 25 = 5^2 \implies x = 2

.

5^x = -5

has no real solution.

An exponential function with a positive base produces strictly positive output values for all real domain inputs.

Key Concept

Solving exponential equations reducible to quadratic form using exponent laws

Question 15Question →

If $x$ is a real number satisfying the exponential equation $4^{x+1} - 4^{x-1} = 120$ , what is the value of $2^{2x + 1}$ ?

16

32

64

128

-64

Show answer & explanation

Answer: 64

Answer

The value of

2^{2x + 1}

is 64.

Factoring

4^{x-1}

from

4^{x+1} - 4^{x-1}

gives

4^{x-1}(16 - 1) = 15 \cdot 4^{x-1} = 120

. Dividing by 15 yields

4^{x-1} = 8

. Rewriting with base 2 gives

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

, so

2x - 2 = 3

, meaning

2x = 5

. Substituting

2x = 5

into

2^{2x+1}

gives

2^{5+1} = 2^6 = 64

.

Step-by-Step Solution

1

Factor out

4^{x-1}

from the left side of the equation

4^{x+1} - 4^{x-1} = 120

.

4^{x-1}(4^2 - 1) = 120

, which simplifies to

4^{x-1}(15) = 120

.

Factoring out the lowest power of 4 allows simplification of the terms on the left side.

2

Divide both sides by 15 to isolate

4^{x-1}

.

4^{x-1} = \frac{120}{15} = 8

.

Isolating the exponential expression is necessary to solve for

x

.

3

Express both sides with a common base of 2.

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

.

Converting to a common prime base allows equating the exponents.

4

Equate exponents to solve for

2x

.

2x - 2 = 3 \implies 2x = 5

.

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

5

Substitute

2x = 5

into the target expression

2^{2x + 1}

.

2^{5 + 1} = 2^6 = 64

.

Evaluating the exact expression requested in the stem.

Key Concept

Factoring exponential expressions with variable exponents and converting bases

Estimated Time:1m 30s

Question 16Question →

If $x$ is a real number that satisfies the equation $x - \sqrt{x + 2} = 4$ , what is the value of $(x + 2)^{\frac{3}{2}}$ ?

27

8

7\sqrt{7} + 2\sqrt{2}

-27

81

Show answer & explanation

Answer: $27$

Answer

27

Isolating the radical gives

x - 4 = \sqrt{x + 2}

. Squaring both sides yields

(x - 4)^2 = x + 2

, which expands to

x^2 - 8x + 16 = x + 2

, or

x^2 - 9x + 14 = 0

. Factoring yields

(x - 7)(x - 2) = 0

, giving solutions

x = 7

and

x = 2

. Testing

x = 7

in the original equation gives

7 - \sqrt{9} = 4

, which is valid. Testing

x = 2

gives

2 - \sqrt{4} = 0 \neq 4

, which is extraneous. Substituting the valid root

x = 7

into

(x + 2)^{\frac{3}{2}}

gives

(7 + 2)^{\frac{3}{2}} = 9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27

.

Step-by-Step Solution

1

Isolate the radical term in the given equation.

x - 4 = \sqrt{x + 2}

Isolating the radical allows both sides to be squared cleanly.

2

Square both sides to eliminate the square root and form a quadratic equation.

(x - 4)^2 = x + 2 \implies x^2 - 8x + 16 = x + 2 \implies x^2 - 9x + 14 = 0

Squaring eliminates the radical and allows standard quadratic solving techniques.

3

Factor the quadratic equation to find candidate solutions.

(x - 7)(x - 2) = 0 \implies x = 7 \text{ or } x = 2

Factoring provides potential real roots.

4

Check candidate solutions in the original equation to eliminate extraneous roots.

For

x = 7

:

7 - \sqrt{7 + 2} = 7 - 3 = 4

(valid).
For

x = 2

:

2 - \sqrt{2 + 2} = 2 - 2 = 0 \neq 4

(extraneous).

Squaring both sides can introduce false solutions that fail the original radical equation.

5

Evaluate the target expression

(x + 2)^{\frac{3}{2}}

using

x = 7

.

(7 + 2)^{\frac{3}{2}} = 9^{\frac{3}{2}} = (9^{\frac{1}{2}})^3 = 3^3 = 27

Applying exponent rules (

a^{\frac{m}{n}} = (\sqrt[n]{a})^m

) gives the exact required value.

Key Concept

Solving radical equations requires isolating the radical, squaring both sides, checking for extraneous solutions introduced by squaring, and evaluating fractional exponents via roots and integer powers.

Estimated Time:2m 0s

Question 17Question →

If $x > 0$ and $x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3$ , what is the value of $x^2 + x^{-2}$ ?

7

47

49

51

81

Show answer & explanation

Answer: 47

Answer

47

Squaring both sides of

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

gives

x + 2 + x^{-1} = 9

, which simplifies to

x + x^{-1} = 7

. Squaring both sides of

x + x^{-1} = 7

gives

x^2 + 2 + x^{-2} = 49

, which yields

x^2 + x^{-2} = 47

.

Step-by-Step Solution

1

Square both sides of the given equation

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

.

(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^2 = 3^2 \implies x + 2(x^{\frac{1}{2}})(x^{-\frac{1}{2}}) + x^{-1} = 9

Applying the binomial expansion identity

(a+b)^2 = a^2 + 2ab + b^2

.

2

Simplify the middle term and solve for

x + x^{-1}

.

x + 2(1) + x^{-1} = 9 \implies x + x^{-1} = 7

Since

x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = x^0 = 1

, subtracting 2 from both sides isolates

x + x^{-1}

.

3

Square both sides of

x + x^{-1} = 7

.

(x + x^{-1})^2 = 7^2 \implies x^2 + 2(x)(x^{-1}) + x^{-2} = 49

Squaring

x + x^{-1}

generates the terms

x^2

and

x^{-2}

.

4

Simplify the middle term and solve for

x^2 + x^{-2}

.

x^2 + 2 + x^{-2} = 49 \implies x^2 + x^{-2} = 47

Subtracting 2 from both sides isolates the desired expression

x^2 + x^{-2}

.

Key Concept

Algebraic Exponents and Binomial Expansion

Question 18Question →

If $x$ is a real number that satisfies the equation $\sqrt{x + 6\sqrt{x - 9}} + \sqrt{x - 6\sqrt{x - 9}} = 10$ , what is the value of $x$ ?

Show answer & explanation

Answer: 34

Answer

34

Using the substitution

u = \sqrt{x - 9} \ge 0

, we have

x = u^2 + 9

. The expressions under the square roots become

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

and

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

. Taking square roots gives

\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10

. For

u \ge 3

, this simplifies to

(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5

. Finally, substituting

u = 5

back yields

x = 5^2 + 9 = 34

.

Step-by-Step Solution

1

Define a variable substitution to simplify the nested radical structure.

Let

u = \sqrt{x - 9}

where

u \ge 0

. Squaring both sides gives

u^2 = x - 9

, so

x = u^2 + 9

.

This substitution allows the expressions inside the outer square roots to be rewritten as polynomials in terms of

u

.

2

Rewrite the expressions under each square root as perfect square trinomials.

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

and

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

.

Expressing terms as perfect squares allows the outer radicals to be simplified.

3

Simplify the square root expressions using absolute values.

\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10

.

For any real number

a

,

\sqrt{a^2} = |a|

. Since

u \ge 0

,

u + 3 > 0

, so

|u + 3| = u + 3

.

4

Solve the absolute value equation across valid domain intervals.

If

u \ge 3

,

|u - 3| = u - 3

, giving

(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5

. If

0 \le u < 3

,

|u - 3| = 3 - u

, giving

(u + 3) + (3 - u) = 6 \neq 10

(no solution). Thus,

u = 5

.

Splitting into cases based on the definition of absolute value isolates the valid root.

5

Substitute

u = 5

back into the expression for

x

.

x = 5^2 + 9 = 25 + 9 = 34

.

Converting from

u

back to

x

provides the solution to the original equation.

Key Concept

Simplifying nested radicals by completing the square under the radical sign and applying the identity

\sqrt{a^2} = |a|

.

Question 19Question →

If $x$ and $y$ are positive integers satisfying $3^x - 3^y = 702$ and $\sqrt{x + y} = 3$ , what is the value of $x^2 - y^2$ ?

9

18

27

45

81

Show answer & explanation

Answer: $27$

Answer

The correct value of

x^2 - y^2

is 27.

Squaring

\sqrt{x + y} = 3

gives

x + y = 9

. Factoring

3^x - 3^y = 702

gives

3^y(3^{x-y} - 1) = 702

. Since

702 = 3^3 \times 26

and

(3^{x-y} - 1)

is coprime to 3, we deduce

3^y = 3^3 \implies y = 3

, and

3^{x-3} - 1 = 26 \implies 3^{x-3} = 27 \implies x = 6

. Substituting

x = 6

and

y = 3

into

x^2 - y^2

gives

36 - 9 = 27

.

Step-by-Step Solution

1

Eliminate the radical from the given linear equation.

Squaring both sides of

\sqrt{x + y} = 3

gives

x + y = 9

.

Squaring both sides removes the square root operator to establish a linear relationship between

x

and

y

.

2

Factor out the common exponential term

3^y

from

3^x - 3^y = 702

.

3^y(3^{x-y} - 1) = 702

.

Since

x

and

y

are positive integers and

702 > 0

, it must be true that

x > y

, allowing factoring by exponent rules

3^x = 3^y \cdot 3^{x-y}

.

3

Find the prime factorization of 702 and match the power of 3.

702 = 27 \times 26 = 3^3 \times 26

, so

3^y(3^{x-y} - 1) = 3^3 \times 26

.

The factor

(3^{x-y} - 1)

is not divisible by 3 because

3^{x-y}

is a multiple of 3 for

x > y

. Therefore, all powers of 3 in 702 must belong to

3^y

.

4

Solve for the values of

y

and

x

.

y = 3

and

3^{x-3} - 1 = 26 \implies 3^{x-3} = 27 = 3^3 \implies x - 3 = 3 \implies x = 6

.

Equating prime component bases yields

y = 3

and

x = 6

, which satisfies

x + y = 6 + 3 = 9

.

5

Calculate the target expression

x^2 - y^2

.

x^2 - y^2 = 6^2 - 3^2 = 36 - 9 = 27

.

Substituting

x = 6

and

y = 3

into

x^2 - y^2

yields 27 (or using

(x+y)(x-y) = 9 \times 3 = 27

).

Key Concept

Factoring exponential expressions using prime factorization, radical simplification, and difference of squares.

Question 20Question →

For all real numbers $x > 0$ , which of the following expressions are equivalent to $\left(\frac{x^{3/2} \cdot \sqrt[3]{x}}{x^{1/6}}\right)^2$ ? Select all such expressions.

Select all that apply

\sqrt[3]{x^{10}}

x^3 \sqrt[3]{x}

x^3 \sqrt{x}

x^5

\left(\sqrt[3]{x^5}\right)^2

Show answer & explanation

Answer: $\sqrt[3]{x^{10}}$ ; $x^3 \sqrt[3]{x}$ ; $\left(\sqrt[3]{x^5}\right)^2$

Answer

The expressions equivalent to the given expression are

\sqrt[3]{x^{10}}

,

x^3 \sqrt[3]{x}

, and

\left(\sqrt[3]{x^5}\right)^2

.

Simplifying the given expression by converting all radical forms to fractional exponents yields

x^{10/3}

. The three expressions

\sqrt[3]{x^{10}}

,

x^3 \sqrt[3]{x}

, and

\left(\sqrt[3]{x^5}\right)^2

each rewrite to

x^{10/3}

when evaluated using standard power laws.

Step-by-Step Solution

1

Convert radical expressions to fractional exponent form inside the parentheses.

\sqrt[3]{x} = x^{1/3}

, so the numerator becomes

x^{3/2} \cdot x^{1/3}

.

Converting all terms to exponent notation allows applying standard exponent addition and subtraction rules.

2

Simplify the numerator by adding exponents.

x^{3/2 + 1/3} = x^{9/6 + 2/6} = x^{11/6}

.

When multiplying exponential terms with the same base, add their exponents using a common denominator.

3

Divide by the denominator by subtracting exponents.

\frac{x^{11/6}}{x^{1/6}} = x^{11/6 - 1/6} = x^{10/6} = x^{5/3}

.

When dividing exponential terms with the same base, subtract the denominator's exponent from the numerator's exponent.

4

Apply the outer exponent of 2.

\left(x^{5/3}\right)^2 = x^{(5/3) \cdot 2} = x^{10/3}

.

When raising a power to another power, multiply the inner and outer exponents.

5

Compare

x^{10/3}

to each option.

\sqrt[3]{x^{10}} = x^{10/3}

,

x^3 \sqrt[3]{x} = x^{3 + 1/3} = x^{10/3}

, and

\left(\sqrt[3]{x^5}\right)^2 = (x^{5/3})^2 = x^{10/3}

are all equivalent.

Matching each candidate expression in fractional exponent form identifies all equivalent choices.

Key Concept

Simplification of algebraic expressions using rules of fractional exponents and radicals

Estimated Time:1m 30s

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