Exponents, Powers, and Square Roots Practice Questions — GRE General Test — Page 2

If

x

is a positive integer such that

\frac{4^x + 4^x + 4^x + 4^x}{2^x + 2^x} = 512

what is the value of

x

?

Show answer & explanation

Answer: 8

Answer

The value of

x

is 8.

Combining four terms of

4^x

yields

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

. Combining two terms of

2^x

yields

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

. Dividing the numerator by the denominator gives

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

. Since

512 = 2^9

, setting

2^{x+1} = 2^9

gives

x + 1 = 9

, which leads directly to

x = 8

.

Step-by-Step Solution

1

Simplify the numerator by combining identical added terms.

The numerator

4^x + 4^x + 4^x + 4^x

equals

4 \cdot 4^x

, which simplifies to

4^{x+1}

.

Adding four identical quantities is equivalent to multiplying that quantity by 4.

2

Simplify the denominator by combining identical added terms.

The denominator

2^x + 2^x

equals

2 \cdot 2^x

, which simplifies to

2^{x+1}

.

Adding two identical quantities is equivalent to multiplying that quantity by 2.

3

Convert the numerator to base 2 and simplify the fraction.

Since

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

, the fraction becomes

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

.

Converting all terms to a common base allows using the exponent quotient rule

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

.

4

Solve for

x

by equating the simplified power to 512.

Setting

2^{x+1} = 512 = 2^9

yields

x + 1 = 9

, so

x = 8

.

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

Key Concept

Combining repeated addition of exponential terms and converting powers to a common base using exponent laws (

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

and

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

).

Question 22Question →

If $x$ is a positive real number such that $x^{\frac{3}{4}} = 27$ , what is the value of $x^{-\frac{1}{2}}$ ?

\frac{1}{9}

9

-\frac{1}{9}

\frac{1}{3}

\frac{1}{27}

Show answer & explanation

Answer: $\frac{1}{9}$

Answer

\frac{1}{9}

Raising both sides of

x^{\frac{3}{4}} = 27

to the power of

\frac{4}{3}

gives

x = (3^3)^{\frac{4}{3}} = 3^4 = 81

. Substituting

x = 81

into

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

gives

81^{-\frac{1}{2}} = \frac{1}{\sqrt{81}} = \frac{1}{9}

.

Step-by-Step Solution

1

Solve for

x

in the equation

x^{\frac{3}{4}} = 27

.

x = 27^{\frac{4}{3}} = (3^3)^{\frac{4}{3}} = 3^4 = 81

Raise both sides to the power of

\frac{4}{3}

to isolate

x

.

2

Evaluate

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

for

x = 81

.

81^{-\frac{1}{2}} = \frac{1}{81^{\frac{1}{2}}} = \frac{1}{\sqrt{81}} = \frac{1}{9}

Apply the negative exponent rule

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

and the fractional exponent rule

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

.

Key Concept

Fractional and Negative Exponents

Estimated Time:1m 30s

Question 23Question →

If $x$ and $y$ are real numbers such that $x < 0 < y$ and $x^2 > y^2$ , which of the following statements must be true? Select all such statements.

Select all that apply

x^3 < y^3

\sqrt{x^2} + x = 0

x + y < 0

\sqrt{x^2 + y^2} = |x| + y

(-x)^2 = -x^2

Show answer & explanation

Answer: $x^3 < y^3$ ; $\sqrt{x^2} + x = 0$ ; $x + y < 0$

Answer

The correct statements are

x^3 < y^3

,

\sqrt{x^2} + x = 0

, and

x + y < 0

.

The statement

x^3 < y^3

is correct because cubing a negative number keeps it negative while cubing a positive number keeps it positive. The statement

\sqrt{x^2} + x = 0

is correct because

\sqrt{x^2} = |x| = -x

for negative numbers. The statement

x + y < 0

is correct because

x^2 > y^2

implies

|x| > y

, meaning the negative component

x

has a larger absolute magnitude than the positive component

y

.

Step-by-Step Solution

1

Analyze the signs of odd powers for

x < 0 < y

x^3 < 0

and

y^3 > 0

, which guarantees

x^3 < y^3

.

Odd powers preserve the original sign of the base.

2

Apply the definition of principal square roots to negative values

\sqrt{x^2} = |x| = -x

, so

\sqrt{x^2} + x = -x + x = 0

.

The square root symbol

\sqrt{}

denotes the principal (non-negative) root.

3

Compare absolute values using

x^2 > y^2

|x| > y \implies -x > y \implies x + y < 0

.

Since

x < 0

, its magnitude

|x|

is

-x

, which dominates the positive value

y

.

4

Evaluate the false options against exponent and radical rules

\sqrt{x^2 + y^2} \neq |x| + y

due to non-distributivity of roots, and

(-x)^2 = x^2 \neq -x^2

.

Radicals do not distribute over sums, and even powers eliminate negative signs.

Key Concept

Properties of real exponents, radical expressions, and absolute values for negative bases

Estimated Time:1m 30s

Question 24Question →

If $x = 4$ , what is the value of $\sqrt{9^x + 9^x + 9^x + 9^x}$ ?

81

162

324

648

1,296

Show answer & explanation

Answer: 162

Answer

162

Combining the four identical terms under the square root gives

\sqrt{4 \cdot 9^4}

. Splitting the root using product rules yields

\sqrt{4} \cdot \sqrt{9^4} = 2 \cdot 9^2 = 2 \cdot 81 = 162

.

Step-by-Step Solution

1

Substitute

x = 4

into the given radical expression.

The expression becomes

\sqrt{9^4 + 9^4 + 9^4 + 9^4}

.

Direct substitution of the given variable value.

2

Combine the four identical terms under the radical.

9^4 + 9^4 + 9^4 + 9^4 = 4 \cdot 9^4

.

Repeated addition of

4

identical terms is equivalent to multiplication by

4

.

3

Apply the product rule for square roots,

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

.

\sqrt{4 \cdot 9^4} = \sqrt{4} \cdot \sqrt{9^4} = 2 \cdot 9^2.

Both

4

and

9^4

are perfect squares.

4

Evaluate the numerical value.

2 \cdot 81 = 162.

Simplifying the arithmetic expression.

Key Concept

Combining like terms under radical sign and applying product rules of square roots

Estimated Time:1m 15s

Question 25Question →

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

Select all that apply

x^2 > y^2

\sqrt{x^2 y^4} = -x y^2

x^3 > y^3

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

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

Show answer & explanation

Answer: $x^2 > y^2$ ; $\sqrt{x^2 y^4} = -x y^2$

Answer

The correct statements are the inequality asserting that the square of the first variable is greater than the square of the second, and the identity simplifying the square root of the product of the powers to negative the product of the first variable and the square of the second.

The statement comparing squared values is correct because any real number less than -1 has an absolute value greater than 1, so its square is strictly greater than 1, while any positive number less than 1 has a square strictly less than 1. The radical identity statement is correct because taking the square root of

x^2

yields

|x|

, which evaluates to

-x

when

x

is negative.

Step-by-Step Solution

1

Analyze the given bounds for both variables.

For the first variable,

x < -1

, which implies

|x| > 1

,

x

is negative,

x^2 > 1

, and

x^3 < -1

. For the second variable,

0 < y < 1

, which implies

y > 0

,

y^2 < 1

, and

y^3 > 0

.

Establishing explicit bounds on magnitudes and signs is necessary to evaluate powers and absolute value roots.

2

Evaluate the inequality comparing the squared terms.

Since

x^2 > 1

and

y^2 < 1

, it follows directly that

x^2 > y^2

.

Transitive comparison across the threshold value of 1 proves the inequality holds.

3

Simplify the radical expression

\sqrt{x^2 y^4}

.

\sqrt{x^2 y^4} = \sqrt{x^2} \cdot \sqrt{y^4} = |x| \cdot y^2

. Since

x < 0

,

|x| = -x

, so the expression simplifies to

-x y^2

.

The principal square root of

x^2

must equal the absolute value

|x|

, which requires a sign flip when

x

is negative.

4

Verify remaining candidate expressions for potential fallacies.

Odd powers preserve negative signs so

x^3 < y^3

;

\sqrt{x^2}

equals

-x \neq x

; and expanding

(x+y)^2

produces a nonzero cross-term

2xy

.

Eliminating false options confirms that only two statements are universally true.

Key Concept

Principal square roots and even/odd power behaviors under negative variable bounds

Question 26Question →

If $m$ and $n$ are negative integers such that $m < n$ , which of the following statements must be true? Select all such statements.

Select all that apply

2^m < 2^n

\left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n

m^2 < n^2

\sqrt{m^2} = m

2^{m+n} = 2^m + 2^n

Show answer & explanation

Answer: $2^m < 2^n$ ; \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n

Answer

The statements

2^m < 2^n

and \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n must be true.

For the statement involving base

2

, since

2 > 1

, the exponential function is strictly increasing, so

m < n

guarantees

2^m < 2^n

. For the statement involving base

\frac{1}{2}

, since

0 < \frac{1}{2} < 1

, the function is strictly decreasing, meaning a smaller input

m

produces a larger output, so \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n.

Step-by-Step Solution

1

Analyze the expression

2^m < 2^n

for base greater than 1

Since

b = 2 > 1

, raising

2

to a larger exponent yields a larger value. Because

m < n

,

2^m < 2^n

is always true.

Exponential functions with a base

b > 1

are strictly increasing.

2

Analyze the expression \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n for fractional base between 0 and 1

Since

b = \frac{1}{2}

is between

0

and

1

, raising

\frac{1}{2}

to a smaller exponent yields a larger value. Because

m < n

, \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n is always true.

Exponential functions with a base

0 < b < 1

are strictly decreasing.

3

Evaluate the remaining algebraic statements using counterexamples

For

m = -3

and

n = -2

:

m^2 = 9 > 4 = n^2

, so

m^2 < n^2

is false.

\sqrt{m^2} = \sqrt{9} = 3 \neq -3

, so

\sqrt{m^2} = m

is false.

2^{-5} = \frac{1}{32} \neq \frac{1}{8} + \frac{1}{4} = \frac{3}{8}

, so

2^{m+n} = 2^m + 2^n

is false.

A single counterexample disproves that a statement MUST be true.

Key Concept

Monotonicity of exponential functions and properties of square roots of negative bases

Question 27Question →

If $x$ is a real number such that $\frac{4^{x+2} - 4^x}{15} = 2^{x+3}$ , what is the value of $x$ ?

1

2

3

5

6

Show answer & explanation

Answer: 3

Answer

3

Factoring out

4^x

from the numerator gives

4^x(4^2 - 1) = 15 \cdot 4^x

. Dividing by 15 simplifies the left-hand side to

4^x

. Expressing

4^x

as

2^{2x}

allows setting

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

. Equating the exponents

2x = x + 3

yields

x = 3

.

Step-by-Step Solution

1

Factor the numerator of the left-hand side

4^{x+2} - 4^x = 4^x(4^2 - 1) = 4^x(16 - 1) = 15 \cdot 4^x

Factoring out the common exponential term

4^x

simplifies the subtraction.

2

Simplify the fraction on the left-hand side

\frac{15 \cdot 4^x}{15} = 4^x

The factor of 15 in the numerator cancels with 15 in the denominator.

3

Rewrite

4^x

in terms of base 2

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

Both sides must have a common base to equate their exponents.

4

Set the exponential expressions equal and solve for

x

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

Since the bases are equal and positive (base 2), their exponents must be equal.

Key Concept

Factoring common exponential terms and equating exponents with identical bases

Estimated Time:1m 30s

Question 28Question →

If $n$ is a real number such that $\frac{27^n + 27^n + 27^n}{3^{n+2}} = 243$ , what is the value of $n$ ?

Show answer & explanation

Answer: 3

Answer

3

Rewriting

27^n + 27^n + 27^n

as

3 \cdot (3^3)^n = 3^{3n+1}

allows the left-hand side to simplify to

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

. Equating this to

243 = 3^5

gives

2n - 1 = 5

, which solves to

n = 3

.

Step-by-Step Solution

1

Express repeated addition in the numerator as multiplication.

27^n + 27^n + 27^n = 3 \cdot 27^n

Adding three identical quantities is equivalent to multiplying one quantity by 3.

2

Convert base 27 to base 3 and apply exponent multiplication.

3 \cdot (3^3)^n = 3^1 \cdot 3^{3n} = 3^{3n+1}

Since

27 = 3^3

, using the power rule

(a^b)^c = a^{bc}

and product rule

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

converts the numerator to a single power of 3.

3

Simplify the fraction using the quotient rule of exponents.

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

Dividing exponential terms with the same base requires subtracting the exponent in the denominator from the exponent in the numerator.

4

Rewrite 243 with base 3 and equate exponents across the equal sign.

3^{2n-1} = 3^5 \implies 2n - 1 = 5

Since

243 = 3^5

, two exponential expressions with the same base are equal if and only if their exponents are equal.

5

Solve the linear equation for

n

.

2n = 6 \implies n = 3

Adding 1 to both sides yields

2n = 6

, and dividing by 2 yields

n = 3

.

Key Concept

Combining repeated addition of exponential terms and converting expressions to a common base using exponent rules.

Exponents, Powers, and Square Roots

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes