Exponents, Powers, and Square Roots Practice Questions — GRE General Test

If $x > 1$ and $\sqrt{x^{\sqrt{x}}} = x^2$ , what is the value of $x$ ?

2

4

8

16

64

Show answer & explanation

Answer: $16$

Answer

16

By rewriting the square root as an exponent of

1/2

, the left side becomes

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

. Since the base

x

is greater than 1, set the exponents equal to each other:

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

, which gives

\sqrt{x} = 4

. Squaring both sides yields

x = 16

.

Step-by-Step Solution

1

Express the radical on the left side of the equation as a fractional exponent.

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

The square root rule states that

\sqrt[k]{a} = a^{\frac{1}{k}}

.

2

Set the exponent of the left side equal to the exponent of the right side.

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

Since the bases are equal (

x > 1

), their corresponding exponents must be equal.

3

Solve for

\sqrt{x}

by multiplying both sides by 2.

\sqrt{x} = 4

Isolating the radical term allows determination of the root value.

4

Square both sides to solve for

x

.

x = 4^2 = 16

Squaring a principal square root yields the underlying radicand.

Key Concept

Combining Fractional Exponents and Radical Expressions

Question 2Question →

If $x$ is a real number such that $\sqrt[4]{16^{x+3}} = 8^{x-1}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The value of

x

is

3

.

To solve

\sqrt[4]{16^{x+3}} = 8^{x-1}

, express both sides with the base 2. The left side simplifies to

\sqrt[4]{(2^4)^{x+3}} = 2^{\frac{4(x+3)}{4}} = 2^{x+3}

. The right side simplifies to

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

. Equating the exponents yields

x + 3 = 3x - 3

, which solves to

2x = 6

, giving

x = 3

.

Step-by-Step Solution

1

Rewrite 16 and 8 using prime base 2

16 = 2^4

and

8 = 2^3

Converting terms to a common base allows direct comparison of exponents.

2

Simplify the left-hand side radical expression

\sqrt[4]{16^{x+3}} = \sqrt[4]{(2^4)^{x+3}} = 2^{\frac{4(x+3)}{4}} = 2^{x+3}

The

n

-th root

\sqrt[n]{a^m}

is equivalent to

a^{m/n}

.

3

Simplify the right-hand side exponential expression

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

Applying the exponent power rule

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

requires multiplying

3

by

(x - 1)

.

4

Equate the exponents and solve for

x

x + 3 = 3x - 3 \implies 2x = 6 \implies x = 3

When

a^u = a^v

for

a > 0

and

a \neq 1

, it follows that

u = v

.

Key Concept

Solving exponential equations using prime base factorization and radical conversion rules

Question 3Question →

If

x

is a positive integer such that

\sqrt{4^{x+1} \cdot 5^{2x} + 4^x \cdot 5^{2x+1}} = 30,000

what is the value of

x

?

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

Factoring the common exponential term

4^x \cdot 5^{2x}

inside the radical yields

\sqrt{(4^x \cdot 5^{2x})(4 + 5)} = \sqrt{9 \cdot 4^x \cdot (5^2)^x} = \sqrt{9 \cdot (4 \cdot 25)^x} = \sqrt{9 \cdot 100^x} = \sqrt{9 \cdot 10^{2x}} = 3 \cdot 10^x

. Setting

3 \cdot 10^x = 30,000

gives

10^x = 10,000 = 10^4

, which means

x = 4

.

Step-by-Step Solution

1

Separate the addition in exponents using exponent rules.

4^{x+1} \cdot 5^{2x} = 4^x \cdot 4^1 \cdot 5^{2x}

and

4^x \cdot 5^{2x+1} = 4^x \cdot 5^{2x} \cdot 5^1

.

Applying the product rule of exponents

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

prepares terms for factoring.

2

Factor out the common expression

4^x \cdot 5^{2x}

from the sum inside the radical.

4^{x+1} \cdot 5^{2x} + 4^x \cdot 5^{2x+1} = (4^x \cdot 5^{2x})(4 + 5) = 9 \cdot 4^x \cdot 5^{2x}

.

Factoring converts the sum under the square root into a single product.

3

Combine terms with powers into base 10.

4^x \cdot 5^{2x} = 4^x \cdot (5^2)^x = 4^x \cdot 25^x = (4 \cdot 25)^x = 100^x = 10^{2x}

.

Using power of a power

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

and power of a product

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

simplifies the expression into powers of 10.

4

Take the square root of the simplified product.

\sqrt{9 \cdot 10^{2x}} = \sqrt{9} \cdot \sqrt{10^{2x}} = 3 \cdot 10^x

.

Applying the product rule for radicals

\sqrt{ab} = \sqrt{a}\sqrt{b}

and halving the exponent

(10^{2x})^{1/2} = 10^x

.

5

Equate the simplified expression to 30,000 and solve for

x

.

3 \cdot 10^x = 30,000 \implies 10^x = 10,000 \implies 10^x = 10^4 \implies x = 4

.

Dividing both sides by 3 isolates

10^x

, and matching exponential bases gives

x = 4

.

Key Concept

Exponent Rules and Radical Simplification

Estimated Time:2m 0s

Question 4Question →

If $x$ and $y$ are positive integers such that $5^x \cdot 2^y = 10^{2x-1} \cdot 4^{x+1}$ , what is the value of $y - x$ ?

2

3

4

5

6

Show answer & explanation

Answer: 4

Answer

The value of

y - x

is 4.

By prime-factorizing the bases on the right-hand side,

10^{2x-1} \cdot 4^{x+1}

becomes

(2 \cdot 5)^{2x-1} \cdot (2^2)^{x+1} = 5^{2x-1} \cdot 2^{4x+1}

. Matching the powers of 5 gives

x = 2x - 1

, which yields

x = 1

. Matching the powers of 2 gives

y = 4x + 1

, which yields

y = 5

. Subtracting

x

from

y

gives

5 - 1 = 4

.

Step-by-Step Solution

1

Rewrite composite bases into prime factor bases on the right side of the equation.

10^{2x-1} = (2 \cdot 5)^{2x-1} = 2^{2x-1} \cdot 5^{2x-1}

and

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

.

Converting all terms to prime bases (2 and 5) allows equating corresponding exponents.

2

Combine terms with identical bases on the right side.

10^{2x-1} \cdot 4^{x+1} = 5^{2x-1} \cdot 2^{(2x-1) + (2x+2)} = 5^{2x-1} \cdot 2^{4x+1}

.

Applying the product rule of exponents

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

simplifies the right side.

3

Equate exponents of corresponding prime bases from both sides of

5^x \cdot 2^y = 5^{2x-1} \cdot 2^{4x+1}

.

Equating powers of 5 yields

x = 2x - 1 \implies x = 1

. Equating powers of 2 yields

y = 4x + 1

.

Since 2 and 5 are distinct prime numbers, their corresponding exponents must be equal.

4

Calculate

y

and evaluate

y - x

.

y = 4(1) + 1 = 5

, so

y - x = 5 - 1 = 4

.

Substituting

x = 1

gives

y = 5

, satisfying the final question requirement.

Key Concept

Decomposing exponential bases into prime factors and applying exponent rules (

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

and

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

) to solve system equations of powers.

Question 5Question →

If

x

is a positive integer such that

\frac{\sqrt{54 \cdot 4^{2x+1} + 9 \cdot 4^{2x}}}{\sqrt{2^{x+3} + 2^x}} = 2560

what is the value of

x

?

Show answer & explanation

Answer: 6

Answer

The value of

x

is

6

.

Factoring out common exponential terms inside both radicals yields

\sqrt{225 \cdot 2^{4x}} = 15 \cdot 2^{2x}

for the numerator and

\sqrt{9 \cdot 2^x} = 3 \cdot 2^{x/2}

for the denominator. Dividing these gives

5 \cdot 2^{3x/2}

. Equating this to

2560

results in

2^{3x/2} = 512 = 2^9

, which simplifies to

\frac{3x}{2} = 9

, or

x = 6

.

Step-by-Step Solution

1

Simplify the numerator inside the radical expression.

\sqrt{54 \cdot 4^{2x+1} + 9 \cdot 4^{2x}} = 15 \cdot 2^{2x}

Rewrite

4^{2x+1}

as

4 \cdot 4^{2x}

. Then factor out

4^{2x}

:

54(4 \cdot 4^{2x}) + 9 \cdot 4^{2x} = (216 + 9)4^{2x} = 225 \cdot 4^{2x}

. Taking the square root gives

\sqrt{225} \cdot \sqrt{(2^2)^{2x}} = 15 \cdot 2^{2x}

.

2

Simplify the denominator inside the radical expression.

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

Rewrite

2^{x+3}

as

2^3 \cdot 2^x = 8 \cdot 2^x

. Factoring out

2^x

gives

(8 + 1)2^x = 9 \cdot 2^x

. Taking the square root gives

\sqrt{9} \cdot \sqrt{2^x} = 3 \cdot 2^{x/2}

.

3

Simplify the quotient of the two radical expressions.

\frac{15 \cdot 2^{2x}}{3 \cdot 2^{x/2}} = 5 \cdot 2^{3x/2}

Divide the constants

\frac{15}{3} = 5

and subtract exponents with the same base:

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

.

4

Equate to 2560 and solve for

x

.

x = 6

Divide both sides by 5:

2^{3x/2} = \frac{2560}{5} = 512

. Express 512 as a power of 2:

512 = 2^9

. Therefore,

\frac{3x}{2} = 9 \implies 3x = 18 \implies x = 6

.

Key Concept

Exponent and Radical Simplification using Base Prime Factorization

Estimated Time:2m 0s

Question 6Question →

If

k

is a positive integer, which of the following expressions is equivalent to

\frac{2^{k+2} \cdot 3^{2k-1} + 6^k \cdot 3^{k-1}}{12^k \cdot 2^{-1}}

for all values of

k

?

\frac{5}{3} \left(\frac{3}{2}\right)^k

\frac{10}{3} \left(\frac{3}{2}\right)^k

\frac{5}{3} \cdot 6^k

5 \left(\frac{3}{2}\right)^k

\frac{10}{3} \cdot 3^k

Show answer & explanation

Answer: $\frac{10}{3} \left(\frac{3}{2}\right)^k$

Answer

The expression simplifies to

\frac{10}{3} \left(\frac{3}{2}\right)^k

.

Factoring the terms in the numerator into base

18^k

gives

\frac{4}{3} \cdot 18^k + \frac{1}{3} \cdot 18^k = \frac{5}{3} \cdot 18^k

. The denominator equals

\frac{1}{2} \cdot 12^k

. Dividing numerator by denominator yields

\frac{5/3}{1/2} \cdot \left(\frac{18}{12}\right)^k = \frac{10}{3} \left(\frac{3}{2}\right)^k

.

Step-by-Step Solution

1

Rewrite each term in the numerator using prime base factorization

2^{k+2} \cdot 3^{2k-1} = 2^k \cdot 2^2 \cdot (3^2)^k \cdot 3^{-1} = 4 \cdot 2^k \cdot 9^k \cdot \frac{1}{3} = \frac{4}{3} \cdot 18^k

, and

6^k \cdot 3^{k-1} = (2 \cdot 3)^k \cdot 3^k \cdot 3^{-1} = 2^k \cdot 9^k \cdot \frac{1}{3} = \frac{1}{3} \cdot 18^k

.

Converting all powers to base 18 allows terms with identical exponential factors to be combined.

2

Combine the terms in the numerator

\frac{4}{3} \cdot 18^k + \frac{1}{3} \cdot 18^k = \left(\frac{4}{3} + \frac{1}{3}\right) \cdot 18^k = \frac{5}{3} \cdot 18^k

.

Adding coefficients of like exponential terms.

3

Simplify the denominator expression

12^k \cdot 2^{-1} = \frac{1}{2} \cdot 12^k

.

Applying the negative exponent rule

a^{-n} = \frac{1}{a^n}

.

4

Divide the numerator by the denominator

\frac{\frac{5}{3} \cdot 18^k}{\frac{1}{2} \cdot 12^k} = \frac{5/3}{1/2} \cdot \left(\frac{18}{12}\right)^k = \left(\frac{5}{3} \cdot 2\right) \cdot \left(\frac{3}{2}\right)^k = \frac{10}{3} \left(\frac{3}{2}\right)^k

Dividing fractions by multiplying by the reciprocal and applying quotient rule for powers with the same exponent.

Key Concept

Prime base factorization and laws of exponents

Question 7Question →

If $x$ is a negative real number such that $\sqrt{(-x)^3 \cdot x^2} = 32$ , what is the value of $x$ ?

-16

-4

-2

2

4

Show answer & explanation

Answer: $-4$

Answer

-4

Simplifying the expression inside the radical gives

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

. Setting

\sqrt{-x^5} = 32

and squaring both sides gives

-x^5 = 32^2 = 1024

, which means

x^5 = -1024

. The fifth root of

-1024

is

-4

. Since

-4

is a negative real number, it satisfies all conditions of the problem.

Step-by-Step Solution

1

Simplify the expression under the square root

Since

(-x)^3 = -x^3

, we have

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

.

Applying exponent addition rules

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

and odd power rules for negative quantities.

2

Square both sides of the equation to eliminate the square root

-x^5 = 32^2 = 1024

.

Squaring both sides of

\sqrt{-x^5} = 32

isolates the radicand.

3

Solve for x

x^5 = -1024 \implies x = \sqrt[5]{-1024} = -4

.

Taking the 5th root of

-1024 = (-2)^{10} = (-4)^5

yields

x = -4

, which satisfies the given condition

x < 0

.

Key Concept

Simplifying expressions with powers and square roots involving negative variables.

Estimated Time:1m 15s

Question 8Question →

If $n$ is a positive integer such that $2^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480$ , what is the value of $n$ ?

Show answer & explanation

Answer: 7

Answer

The value of

n

is 7.

Factoring out

2^{n-2}

converts the sum into

2^{n-2}(1 + 2 + 4 + 8) = 15 \cdot 2^{n-2} = 480

. Dividing 480 by 15 gives

2^{n-2} = 32

. Since

32 = 2^5

, setting

n - 2 = 5

yields

n = 7

.

Step-by-Step Solution

1

Factor out the lowest power of 2, which is

2^{n-2}

, from all terms on the left side of the equation.

2^{n-2}(1 + 2 + 4 + 8) = 480

Factoring out a common exponential term simplifies the addition of powers into a product of a single exponential term and a constant.

2

Evaluate the constant factor inside the parentheses and solve for the exponential expression

2^{n-2}

.

15 \cdot 2^{n-2} = 480 \implies 2^{n-2} = 32

Summing

1 + 2 + 4 + 8

yields 15. Dividing both sides by 15 isolates the base-2 term.

3

Express 32 as a power with base 2 and equate the exponents.

2^{n-2} = 2^5 \implies n - 2 = 5 \implies n = 7

Since

32 = 2^5

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

Key Concept

Factoring sum of exponential terms with common bases

Question 9Question →

If $x$ is a negative real number, which of the following expressions are equivalent to $\sqrt{-x^3}$ ? Select all such expressions.

Select all that apply

-x\sqrt{-x}

x\sqrt{-x}

|x|\sqrt{-x}

(-x)^{3/2}

-x^{3/2}

Show answer & explanation

Answer: $-x\sqrt{-x}$ ; $|x|\sqrt{-x}$ ; $(-x)^{3/2}$

Answer

The expressions equivalent to

\sqrt{-x^3}

are

-x\sqrt{-x}

,

|x|\sqrt{-x}

, and

(-x)^{3/2}

.

Because

x

is negative,

-x

is a positive quantity. We can express

-x^3

as

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

. Taking the principal square root yields

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

. Because

|x| = -x

for negative numbers, the expression

|x|\sqrt{-x}

is identical to

-x\sqrt{-x}

. Furthermore, converting to rational exponents gives

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

. Thus, all three of these expressions are mathematically equivalent to the original radical expression.

Step-by-Step Solution

1

Analyze the sign of the base and inside of the radical

Since

x < 0

, the quantity

-x

is strictly positive (

-x > 0

). Consequently,

-x^3 = (-x)^3 > 0

, ensuring

\sqrt{-x^3}

is a real, non-negative number.

Principal square roots require a non-negative radicand and yield a non-negative result in real arithmetic.

2

Simplify the radical expression using perfect squares

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

.

Because

-x > 0

,

\sqrt{(-x)^2} = -x

.

3

Evaluate equivalent representations using absolute value and rational exponents

Since

x < 0

,

|x| = -x

, so

|x|\sqrt{-x} = -x\sqrt{-x}

. Also,

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

.

Both rewrite rules preserve both magnitude and non-negative sign for all

x < 0

.

Key Concept

Simplifying radicals and fractional exponents with negative variable bases

Question 10Question →

If $2^x \times 4^3 = 2^9$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

By rewriting

4^3

as

(2^2)^3 = 2^6

, the equation becomes

2^x \cdot 2^6 = 2^9

. Applying the product rule gives

2^{x+6} = 2^9

, which simplifies to

x + 6 = 9

, yielding

x = 3

.

Step-by-Step Solution

1

Convert all terms to base 2

4^3 = (2^2)^3 = 2^{2 \times 3} = 2^6

To apply exponent rules, expressions should share a common base.

2

Apply product rule of exponents to the left side of the equation

2^x \times 2^6 = 2^{x+6}

When multiplying exponential terms with the same base, add their exponents:

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

.

3

Equate the exponents from both sides

x + 6 = 9

Since the bases are identical and non-zero, the exponents must be equal.

4

Solve the linear equation for

x

x = 3

Subtract 6 from both sides of the equation.

Key Concept

Combining exponential terms with equal bases using power rules (

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

and

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

).

Question 11Question →

What is the value of $\sqrt{4^5 + 4^5 + 4^5 + 4^5}$ ?

16

64

128

256

1024

Show answer & explanation

Answer: 64

Answer

64

Combining the four terms inside the radical gives

4 \times 4^5 = 4^6

. Taking the square root yields

\sqrt{4^6} = 4^3

, which evaluates to 64.

Step-by-Step Solution

1

Rewrite the sum inside the radical as multiplication

4^5 + 4^5 + 4^5 + 4^5 = 4 \times 4^5

Adding four identical terms is equivalent to multiplying the term by 4.

2

Apply the product rule for exponents

4^1 \times 4^5 = 4^{1+5} = 4^6

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

3

Simplify the radical expression

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

Taking the square root of a power is equivalent to dividing the exponent by 2.

Key Concept

Exponent Addition and Radical Laws

Question 12Question →

What is the value of $\frac{2^7 + 2^7}{2^4}$ ?

1

8

16

128

1024

Show answer & explanation

Answer: 16

Answer

16

Combining the numerator yields

2^7 + 2^7 = 2(2^7) = 2^8

. Dividing

2^8

by

2^4

using the quotient rule gives

2^{8-4} = 2^4 = 16

.

Step-by-Step Solution

1

Simplify the numerator by factoring out common terms

2^7 + 2^7 = 2 \times 2^7 = 2^{1+7} = 2^8

Adding two identical exponential terms is equivalent to multiplying one term by 2.

2

Apply the quotient rule of exponents

\frac{2^8}{2^4} = 2^{8-4} = 2^4

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

3

Evaluate the power

2^4 = 16

Compute the numerical value of 2 raised to the 4th power.

Key Concept

Combining like exponential terms and applying the quotient rule of exponents.

Question 13Question →

If $x$ is a real number such that $x^4 = 16$ , which of the following values could be equal to $x^3$ ? Select all that apply.

Select all that apply

-8

-4

4

8

16

Show answer & explanation

Answer: $-8$ ; $8$

Answer

The values

-8

and

8

are the possible values of

x^3

.

Solving

x^4 = 16

gives two real solutions,

x = 2

and

x = -2

, because raising any real number to an even power yields a non-negative result. Cubing each solution gives

2^3 = 8

and

(-2)^3 = -8

. Thus, both

-8

and

8

are correct values for

x^3

.

Step-by-Step Solution

1

Find all real solutions for

x

in the equation

x^4 = 16

.

x = 2

or

x = -2

.

Taking the fourth root of both sides gives

|x| = \sqrt[4]{16} = 2

, so

x

can be positive or negative.

2

Calculate

x^3

for the positive root

x = 2

.

2^3 = 8

.

Cubing a positive number yields a positive result.

3

Calculate

x^3

for the negative root

x = -2

.

(-2)^3 = -8

.

Cubing a negative number yields a negative result because an odd exponent preserves the sign.

Key Concept

Even and odd power rules for positive and negative real bases

Question 14Question →

If $x$ and $y$ are positive integers such that $2^x - 2^y = 1920$ , what is the value of $x + y$ ?

11

15

18

22

25

Show answer & explanation

Answer: $18$

Answer

The correct value is 18.

Factoring

2^y

from

2^x - 2^y

yields

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

. Factoring 1920 as

2^7 \times 15

allows us to uniquely match the power-of-two factor

2^y = 2^7

(

y = 7

) and the odd factor

2^{x-y} - 1 = 15

(

x - y = 4 \implies x = 11

). Summing

x

and

y

gives

11 + 7 = 18

.

Step-by-Step Solution

1

Factor out the smaller power of 2 from the expression

2^y(2^{x-y} - 1) = 1920

Since

x

and

y

are positive integers and

2^x - 2^y > 0

, we know

x > y

. Factoring

2^y

separates the even power of 2 component from an odd component.

2

Determine the prime factorization of 1920 into a power of 2 and an odd integer

1920 = 128 \times 15 = 2^7 \times 15

Repeatedly dividing 1920 by 2 gives

1920 = 2^7 \times 15

, where 15 is an odd integer.

3

Equate the even and odd components

2^y = 2^7 \implies y = 7

and

2^{x-y} - 1 = 15 \implies 2^{x-y} = 16 = 2^4 \implies x - y = 4

Because

x - y \ge 1

, the quantity

2^{x-y} - 1

must be an odd integer, forcing it to equal 15 and the power of 2 factor to equal

2^7

.

4

Solve for

x

and calculate the final sum

x + y

x = 7 + 4 = 11

, so

x + y = 11 + 7 = 18

Adding

x = 11

and

y = 7

gives the required value.

Key Concept

Factoring exponential expressions by pulling out the common base power and matching unique prime factorizations.

Question 15Question →

If $x$ and $y$ are real numbers such that $-1 < x < 0 < y < 1$ , which of the following statements MUST be true? Select all such statements.

Select all that apply

x^3 < x^5

\sqrt{x^2} = x

\left(y^{1/2}\right)^x > 1

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

x^4 > x^2

Show answer & explanation

Answer: $x^3 < x^5$ ; $\left(y^{1/2}\right)^x > 1$

Answer

The statements

x^3 < x^5

and

\left(y^{1/2}\right)^x > 1

MUST be true.

For

-1 < x < 0 < y < 1

, odd powers of

x

satisfy

-1 < x < x^3 < x^5 < 0

, making the inequality comparing

x^3

and

x^5

true. Furthermore,

y^{1/2}

lies in

(0, 1)

, and raising a base in

(0, 1)

to a negative exponent

x

produces a result strictly greater than 1.

Step-by-Step Solution

1

Analyze the odd power inequality

x^3 < x^5

for

-1 < x < 0

.

Since

x \in (-1, 0)

,

x^2 \in (0, 1)

. Multiplying

-1 < x < 0

by

x^2

gives

x < x^3 < x^5 < 0

. Thus,

x^3 < x^5

holds.

Odd powers preserve negative signs, and higher powers of fractions between 0 and 1 have smaller absolute values.

2

Evaluate the principal square root

\sqrt{x^2}

.

By definition,

\sqrt{x^2} = |x|

. Since

x < 0

,

|x| = -x \neq x

.

The principal square root of a real number is always non-negative.

3

Analyze the expression

\left(y^{1/2}\right)^x

.

Since

0 < y < 1

,

0 < y^{1/2} < 1

. Let

k = y^{1/2}

. Then

k^x = \left(\frac{1}{k}\right)^{-x}

. Since

k < 1

,

\frac{1}{k} > 1

, and since

x < 0

,

-x > 0

. A base greater than 1 raised to a positive power is greater than 1.

Negative exponents indicate the reciprocal of the base.

4

Evaluate

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

.

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

. Because

x < 0

and

y > 0

,

2xy < 0

, so

(x + y)^2 < x^2 + y^2

.

Exponents do not distribute over addition, and cross-terms must be accounted for.

5

Evaluate

x^4 > x^2

.

For

-1 < x < 0

,

0 < x^2 < 1

. Squaring a number in

(0, 1)

yields a smaller number, so

x^4 < x^2

.

Higher even powers of quantities with magnitude less than 1 decrease in magnitude.

Key Concept

Properties of exponents, fractional powers, and principal square roots for bounded negative and positive real numbers

Estimated Time:2m 30s

Question 16Question →

If $3^x + 3^{-x} = 4$ , what is the value of $\frac{27^x + 27^{-x} - 2}{9^x + 9^{-x} + 1}$ ?

\frac{10}{3}

\frac{50}{17}

\frac{62}{17}

\frac{74}{15}

\frac{50}{19}

Show answer & explanation

Answer: $\frac{10}{3}$

Answer

\frac{10}{3}

Squaring

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

yields

9^x + 2 + 9^{-x} = 16

, which simplifies to

9^x + 9^{-x} = 14

. Utilizing the sum of cubes identity gives

27^x + 27^{-x} = (3^x + 3^{-x})(9^x - 1 + 9^{-x}) = 4(14 - 1) = 52

. Substituting these values into the given fraction gives

\frac{52 - 2}{14 + 1} = \frac{50}{15} = \frac{10}{3}

.

Step-by-Step Solution

1

Square the given expression

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

to determine

9^x + 9^{-x}

.

(3^x + 3^{-x})^2 = 9^x + 2(3^x)(3^{-x}) + 9^{-x} = 9^x + 2 + 9^{-x} = 16

, which yields

9^x + 9^{-x} = 14

.

Expanding the square of a binomial requires accounting for the middle term

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

.

2

Express

27^x + 27^{-x}

using the sum of cubes factorization identity.

27^x + 27^{-x} = (3^x)^3 + (3^{-x})^3 = (3^x + 3^{-x})(9^x - 3^x \cdot 3^{-x} + 9^{-x}) = 4(14 - 1) = 52

.

The identity

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

allows factoring cubic exponential expressions.

3

Substitute the evaluated terms into the target rational expression and simplify.

\frac{27^x + 27^{-x} - 2}{9^x + 9^{-x} + 1} = \frac{52 - 2}{14 + 1} = \frac{50}{15} = \frac{10}{3}.

Replacing component expressions with their computed values yields the simplified numerical fraction.

Key Concept

Evaluating high-power exponential expressions using polynomial identity transformations and exponent laws.

Question 17Question →

If $x$ is a real number such that $0 < x < 1$ , which of the following expressions is equivalent to $\sqrt{\frac{x^{-2} - 1}{x^{-2} + 2 + x^2}}$ ?

\frac{\sqrt{1-x^2}}{1+x^2}

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

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

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

\frac{1-x}{1+x}

Show answer & explanation

Answer: $\frac{\sqrt{1-x^2}}{1+x^2}$

Answer

The expression

\frac{\sqrt{1-x^2}}{1+x^2}

is equivalent to the given radical expression.

Expressing

x^{-2}

as

\frac{1}{x^2}

allows the numerator to be rewritten as

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

and the denominator as

\frac{(1+x^2)^2}{x^2}

. Dividing these fractions cancels out

x^2

, leaving

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

under the radical. Taking the square root of the numerator and denominator separately gives

\frac{\sqrt{1-x^2}}{1+x^2}

.

Step-by-Step Solution

1

Rewrite negative exponents as fractions in both the numerator and denominator.

Numerator:

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

. Denominator:

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

.

Converting negative exponents into positive fractional exponents allows common denominators to be established.

2

Factor the perfect square trinomial in the denominator.

The numerator of the denominator expression is

1 + 2x^2 + x^4 = (1 + x^2)^2

. Thus, the entire denominator is

\frac{(1+x^2)^2}{x^2}

.

Recognizing

1 + 2x^2 + x^4

as

(1+x^2)^2

simplifies taking the square root.

3

Simplify the quotient inside the radical.

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

Canceling the common factor of

x^2

in the denominators reduces the nested fraction.

4

Apply the square root rule

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

.

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

Since

1+x^2 > 0

for all real

x

,

\sqrt{(1+x^2)^2} = 1+x^2

.

Key Concept

Simplifying radical expressions containing negative powers and algebraic fractions

Estimated Time:2m 0s

Question 18Question →

If

x

is a real number such that

\sqrt{\frac{25^{x+1} - 25^x}{5^{x+2} + 5^{x+1}}} = 250

what is the value of

x

?

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

Factoring out common powers in the numerator and denominator yields

25^x(25-1) = 24 \cdot 5^{2x}

and

5^{x+1}(5+1) = 6 \cdot 5^{x+1}

. Simplifying their ratio inside the square root gives

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{x-1}

. Taking the square root gives

2 \cdot 5^{(x-1)/2}

. Setting this equal to 250 yields

5^{(x-1)/2} = 125 = 5^3

. Equating the exponents gives

(x-1)/2 = 3

, which solves to

x = 7

.

Step-by-Step Solution

1

Factor the numerator and express terms with a common base of 5

25^{x+1} - 25^x = 25^x(25 - 1) = 24 \cdot (5^2)^x = 24 \cdot 5^{2x}

Factoring out

25^x

simplifies the difference into a single term with base 5.

2

Factor the denominator using base 5

5^{x+2} + 5^{x+1} = 5^{x+1}(5 + 1) = 6 \cdot 5^{x+1}

Factoring out the common power

5^{x+1}

simplifies the sum into a single term.

3

Simplify the fraction inside the square root

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{2x - (x+1)} = 4 \cdot 5^{x-1}

Dividing coefficients (24/6 = 4) and applying exponent rules for division (

a^m / a^n = a^{m-n}

).

4

Take the square root of the simplified expression

\sqrt{4 \cdot 5^{x-1}} = \sqrt{4} \cdot \sqrt{5^{x-1}} = 2 \cdot 5^{\frac{x-1}{2}}

Using radical rules

\sqrt{ab} = \sqrt{a}\sqrt{b}

and

\sqrt{a^k} = a^{k/2}

.

5

Set the simplified radical expression equal to 250 and solve for x

2 \cdot 5^{\frac{x-1}{2}} = 250 \implies 5^{\frac{x-1}{2}} = 125 \implies 5^{\frac{x-1}{2}} = 5^3 \implies \frac{x-1}{2} = 3 \implies x = 7

Dividing both sides by 2 gives

5^{(x-1)/2} = 125 = 5^3

. Equating exponents yields

(x-1)/2 = 3

, so

x = 7

.

Key Concept

Exponent rules, base conversion, factoring exponential terms, and radical simplification

Estimated Time:2m 0s

Question 19Question →

If $x$ and $y$ are positive integers such that $3^{x+2} \cdot 4^y - 3^x \cdot 4^{y+1} = 11,520$ , what is the value of $x + y$ ?

5

6

7

8

10

Show answer & explanation

Answer: 6

Answer

6

Factoring

3^x \cdot 4^y

from the expression

3^{x+2} \cdot 4^y - 3^x \cdot 4^{y+1}

yields

3^x \cdot 4^y (3^2 - 4^1) = 5 \cdot 3^x \cdot 4^y

. Setting this equal to 11,520 and dividing by 5 gives

3^x \cdot 4^y = 2,304

. Prime factorization of 2,304 gives

3^2 \cdot 4^4

, so

x = 2

and

y = 4

. The sum

x + y

is equal to 6.

Step-by-Step Solution

1

Factor out the greatest common exponential factor

3^x \cdot 4^y

from the left side of the equation.

3^x \cdot 4^y (3^2 - 4^1) = 11,520

By exponent rules,

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

and

4^{y+1} = 4^y \cdot 4^1

.

2

Evaluate the constant factor inside the parentheses.

3^2 - 4^1 = 9 - 4 = 5

, so

5 \cdot 3^x \cdot 4^y = 11,520

Simplifying numerical exponents.

3

Divide both sides of the equation by 5.

3^x \cdot 4^y = 2,304

Isolating the variable exponential terms.

4

Determine the prime factorization of 2,304 into powers of 3 and 4.

2,304 = 9 \cdot 256 = 3^2 \cdot 4^4

, which implies

x = 2

and

y = 4

Unique factorization for integer bases.

5

Calculate the sum

x + y

.

x + y = 2 + 4 = 6

Answering the explicit prompt.

Key Concept

Factoring Exponential Expressions and Unique Factorization

Question 20Question →

If $a$ and $b$ are nonzero real numbers such that $a^2 b^3 < 0$ and $\sqrt{a^2 b^4} = -a b^2$ , which of the following statements must be true? Select all such statements.

Select all that apply

a + b < 0

\frac{a^3}{b} > 0

\sqrt{a^4 b^2} = a^2 b

(-a)^3 b^2 < 0

\sqrt{(a + b)^2} = a + b

Show answer & explanation

Answer: $a + b < 0$ ; $\frac{a^3}{b} > 0$

Answer

The statements that must be true are the inequality asserting that the sum of the variables is negative (

a + b < 0

) and the inequality asserting that the quotient of the cubed variable and the second variable is positive (

\frac{a^3}{b} > 0

).

Analyzing the given constraints reveals that both variables are negative. From

a^2 b^3 < 0

, since

a^2 > 0

, we must have

b^3 < 0

, so

b < 0

. Next, from

\sqrt{a^2 b^4} = |a| b^2 = -a b^2

, dividing by

b^2 > 0

gives

|a| = -a

, which implies

a < 0

. Thus,

a < 0

and

b < 0

. The statement asserting

a + b < 0

is true because the sum of two negative numbers is negative. The statement asserting

\frac{a^3}{b} > 0

is true because

a^3 < 0

and

b < 0

, and dividing two negative numbers yields a positive quotient.

Step-by-Step Solution

1

Determine the sign of

b

using the given inequality

a^2 b^3 < 0

.

b < 0

Since

a

is a nonzero real number,

a^2 > 0

. For the product

a^2 b^3

to be negative,

b^3

must be negative, which implies

b < 0

.

2

Determine the sign of

a

using the identity

\sqrt{a^2 b^4} = -a b^2

.

a < 0

Simplify the radical:

\sqrt{a^2 b^4} = \sqrt{a^2} \cdot \sqrt{(b^2)^2} = |a| b^2

. Equating this to

-a b^2

gives

|a| b^2 = -a b^2

. Since

b \neq 0

,

b^2 > 0

, so dividing by

b^2

yields

|a| = -a

. For a nonzero real number,

|a| = -a

implies

a < 0

.

3

Evaluate statement

a + b < 0

.

True

The sum of two negative numbers (

a < 0

and

b < 0

) is always negative.

4

Evaluate statement

\frac{a^3}{b} > 0

.

True

Since

a < 0

,

a^3 < 0

. Dividing the negative quantity

a^3

by the negative quantity

b

yields a positive result.

5

Evaluate statement

\sqrt{a^4 b^2} = a^2 b

.

False

\sqrt{a^4 b^2} = \sqrt{a^4}\sqrt{b^2} = a^2 |b|

. Since

b < 0

,

|b| = -b

, so

\sqrt{a^4 b^2} = -a^2 b

.

6

Evaluate statement

(-a)^3 b^2 < 0

.

False

Since

a < 0

,

-a > 0

, making

(-a)^3 > 0

. Since

b \neq 0

,

b^2 > 0

. The product of two positive numbers is positive, so

(-a)^3 b^2 > 0

.

7

Evaluate statement

\sqrt{(a + b)^2} = a + b

.

False

\sqrt{x^2} = |x|

for any real

x

. Since

a + b < 0

,

\sqrt{(a + b)^2} = |a + b| = -(a + b) \neq a + b

.

Key Concept

Properties of even exponents, odd exponents, and principal square roots of negative variable terms.

Exponents, Powers, and Square Roots

Answer

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