GRE Quantitative Quiz: Exponents Roots

20 questions · exam conditions

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Exponents RootsQuestion 1 of 20

If $x$ is a real number and $\sqrt{x}=\dfrac{1}{3}$ , what is the value of $x^{3/2}$ ?

\dfrac{1}{9}

\dfrac{1}{27}

\dfrac{1}{3}

\dfrac{1}{81}

\dfrac{3}{2}

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GRE Quantitative Quiz

GRE Quantitative Quiz: Exponents Roots

Practice Exponents Roots in GRE Quantitative with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Exponents Roots, giving you a quick way to practice the rules, question types, and explanations that matter most for GRE Quantitative.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If $x$ is a real number and $\sqrt{x}=\dfrac{1}{3}$ , what is the value of $x^{3/2}$ ?

$\dfrac{1}{9}$
$\dfrac{1}{27}$ (correct answer)
$\dfrac{1}{3}$
$\dfrac{1}{81}$
$\dfrac{3}{2}$

Explanation: This question tests composing exponents from root equations. The relevant rule is that (

\sqrt{x}

x^{1/2}

\frac{1}{3}

), so (x = left(\frac{1}{3} ight)^2 = \frac{1}{9}). Then, (x^{3/2} = (x^{1/2})^3 = left(\frac{1}{3} ight)^3 = \frac{1}{27}). Alternatively, (left(\frac{1}{9} ight)^{3/2} = \frac{1}{9} cdot sqrt{\frac{1}{9}} = \frac{1}{9} cdot \frac{1}{3} = \frac{1}{27}). The result is justified by consistent application of exponents. A distractor like choice A, (

\frac{1}{9}

), fails by stopping at (x) instead of applying the 3/2 exponent.

Question 2

If $m$ and $n$ are positive integers such that $2^m=8^n$ , what is the value of $\dfrac{m}{n}$ ?

$\dfrac{1}{3}$
$1$
$3$ (correct answer)
$\dfrac{3}{2}$
$\dfrac{2}{3}$

Explanation: This question tests expressing numbers with the same base to find ratios of exponents. The relevant rule is to rewrite 8 as (

2^3

), so (2^m = (2^3)^n = 2^{3n}). Equating exponents, (m = 3n), so (

\frac{m}{n}

= 3). This holds for positive integers (m) and (n). The result is justified as it satisfies the equation, like (m=3, n=1): (

2^3

8^1

). A distractor like choice A, (

\frac{1}{3}

), fails by inverting the ratio, perhaps from miswriting (8 =

2^{1/3}

Question 3

Which of the following is equivalent to $\dfrac{\sqrt{x^6}}{x^2}$ for real $x\ne 0$ ?

$x$
$x^2$
$|x|$ (correct answer)
$|x|^3$
$\dfrac{1}{|x|}$

Explanation: This question tests simplifying root expressions with absolute values for real numbers. The relevant rule is that (sqrt{x^6} = (x^6)^{1/2} = |x^3| = |x|^3). Dividing by (

x^2

) gives (

|x|^3

x^2

|x|^3

|x|^2

= |x|), since (

x^2

|x|^2

). This holds for all real (x eq 0). The result is justified by checks like (x = -2): (

\sqrt{64}

/4 = 8/4 = 2 = |-2|). A distractor like choice A, (x), fails for negative (x), where it would give a negative instead of positive.

Question 4

Which of the following expressions is equivalent to $\left(16x^4\right)^{\frac{1}{2}}$ for real $x$ ?

$2x^2$
$4|x|$
$|4x^2|$
$4x^2$ (correct answer)
$4x$

Explanation: This question tests the power rule for exponents and understanding of even roots. The rule states that

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

. Applying this:

(16x^4)^{1/2} = 16^{1/2} \cdot (x^4)^{1/2} = 4 \cdot x^{4 \cdot 1/2} = 4x^2

. Note that

x^2

is always non-negative for real

x

, so the expression

4x^2

is well-defined. We can verify: when

x = 2

(16 \cdot 2^4)^{1/2} = (16 \cdot 16)^{1/2} = 256^{1/2} = 16

, and

4 \cdot 2^2 = 4 \cdot 4 = 16

✓. A common mistake would be to think

(x^4)^{1/2} = |x^2|

2|x|

, not recognizing that

x^2

is already non-negative.

Question 5

Which of the following is true for all real numbers $t$ such that $t\neq 0$ ?

$\left(t^2\right)^{\frac{1}{2}}=t$
$\left(t^3\right)^{\frac{1}{3}}=|t|$
$\left(t^4\right)^{\frac{1}{2}}=t^2$ (correct answer)
$\left(t^2\right)^{\frac{3}{2}}=t^3$
$\left(t\right)^{\frac{1}{2}}=\sqrt{|t|}$

Explanation: This question tests understanding of how fractional exponents behave with different bases. We need to check each option for all real

t \neq 0

. Option A:

(t^2)^{1/2} = |t|

, not

t

, since

\sqrt{t^2} = |t|

. Option B:

(t^3)^{1/3} = t

, not

|t|

, since cube roots preserve sign. Option C:

(t^4)^{1/2} = t^2

is correct because

(t^4)^{1/2} = t^{4 \cdot 1/2} = t^2

, and this works for all real

t

. Option D:

(t^2)^{3/2} = |t|^3

, not

t^3

, when

t < 0

. Option E:

t^{1/2}

is undefined for

t < 0

. Therefore, only option C is true for all real

t \neq 0

Question 6

Which of the following expressions is equivalent to $\sqrt[3]{a^2}\,\sqrt[3]{a}$ for real $a$ ?

$\sqrt[3]{a^3}$
$a^2$
$a$ (correct answer)
$\sqrt[3]{a^2}$
$\sqrt[3]{a}$

Explanation: This question tests the multiplication rule for radicals with the same index. The rule states that

\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}

for real numbers where the radicals are defined. Applying this rule:

\sqrt[3]{a^2} \cdot \sqrt[3]{a} = \sqrt[3]{a^2 \cdot a} = \sqrt[3]{a^3}

. Since

\sqrt[3]{a^3} = a

for all real

a

(the cube root function is defined for all real numbers and preserves sign), the answer is

a

. A common mistake would be to add the exponents incorrectly, thinking

\sqrt[3]{a^2} \cdot \sqrt[3]{a} = \sqrt[3]{a^{2+1/3}}

, which misunderstands how radical multiplication works.

Question 7

Which of the following is equivalent to $\dfrac{2^{5/2}}{2^{1/2}}$ ?

$2^2$ (correct answer)
$2^3$
$2^{5}$
$2^{2/2}$
$\sqrt{2^4}$

Explanation: This question tests subtracting exponents with the same base. The relevant rule is that (

\frac{2^{5/2}}{2^{1/2}}

2^{5/2 - 1/2}

2^{4/2}

2^2

= 4). This applies the quotient rule for exponents. The result equals 4, matching (

2^2

). The justification is the exponent arithmetic simplifying correctly. A distractor like choice D, (

2^{2/2}

2^1

= 2), fails by misapplying the fraction in the exponent.

Question 8

Which of the following is equal to $\dfrac{\sqrt{18}}{\sqrt{2}}$ ?

$\sqrt{9}$
$\sqrt{16}$
$3$ (correct answer)
$9$
$6$

Explanation: This question tests simplifying ratios of square roots. The relevant rule is that (

\frac{sqrt{18}}{sqrt{2}}

\sqrt{\frac{18}{2}}

\sqrt{9}

). Simplifying further, (

\sqrt{9}

= 3). Alternatively, factor as (

\frac{3sqrt{2}}{sqrt{2}}

= 3). The result is justified as it eliminates the roots correctly. A distractor like choice D, 9, fails by incorrectly squaring the entire expression or misapplying exponent rules.

Question 9

Which of the following is equivalent to $\left(\dfrac{1}{16}\right)^{3/4}$ ?

$\dfrac{1}{8}$ (correct answer)
$\dfrac{1}{4}$
$\dfrac{1}{2}$
$\dfrac{1}{16}$
$8$

Explanation: This question tests evaluating fractional exponents on fractions. The relevant rule is that (left(\frac{1}{16} ight)^{3/4} = (16^{-1})^{3/4} = 16^{-3/4}). Since (16 =

2^4

), (16^{-3/4} = (2^4)^{-3/4} = 2^{-3} = \frac{1}{8}). Alternatively, (left(\frac{1}{16} ight)^{1/4} = \frac{1}{2}), then raised to the third power is (left(\frac{1}{2} ight)^3 = \frac{1}{8}). The result is justified by consistent exponent simplification. A distractor like choice B, (

\frac{1}{4}

), fails by using an incorrect exponent like 1/2 instead of 3/4.

Question 10

Which of the following is true for all real numbers $x$ such that the expressions are defined?

$\sqrt{x^2} = \ ?$

$x$
$|x|$ (correct answer)
$\pm x$
$x^2$
$\dfrac{1}{|x|}$

Explanation: This question tests your understanding of the relationship between square roots and absolute values. The key principle is that

\sqrt{x^2} = |x|

for all real numbers

x

. This is because squaring any real number (positive or negative) gives a non-negative result, and the square root function returns the non-negative value. For example,

\sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|

. The expression equals the absolute value of

x

, not just

x

itself, because when

x

is negative,

\sqrt{x^2}

gives the positive value. Choice A (

x

) would be incorrect for negative values of

x

, as it doesn't account for the sign change.

Question 11

If $x^{3/2}=27$ , what is $x$ ?

3
9 (correct answer)
18
81

Explanation: Raise both sides to the power 2/3 to undo the exponent 3/2. Then x = 27^(2/3). Take the cube root of 27 first, which is 3, then square it: 3^2 = 9. A tempting wrong answer is 3, but that treats 27 as x^3 instead of x^(3/2), so it doesn't satisfy the equation.

Question 12

If $2^{x^2-2x}=8$ , what is the sum of all possible values of $x$ ?

4
3
-1
2 (correct answer)

Explanation: Rewrite 8 as 2^3, so the exponents must match: x^2 - 2x = 3. Bring all terms to one side: x^2 - 2x - 3 = 0, which factors as (x - 3)(x + 1) = 0, giving x = 3 and x = -1. Their sum is 3 + (-1) = 2. The tempting wrong answer is 3, but that's only one of the two possible values, not the sum.

Question 13

If $4^x=5$ , what is $2^{-2x}$ ?

1/5 (correct answer)
-5
1/25
25

Explanation: Rewrite 4^x as (2^2)^x = 2^(2x). Since 4^x = 5, you have 2^(2x) = 5, and 2^(-2x) is the reciprocal, so the value is 1/5. The tempting wrong answer is 1/25, which comes from mistakenly using exponent -4x instead of -2x.

Question 14

If $n<0$ , which equals $\sqrt[3]{n^6}$ ?

$n^2$ (correct answer)
$-n^2$
$n^3$
$-n^3$

Explanation: Since n^6 = (n^2)^3, taking its cube root gives n^2. The negative condition doesn't change this: n^2 is positive, as the cube root of a positive number must be. Don't pick -n^3; although it is positive when n<0, it isn't equal to n^2 (try n=-2: 8 vs 4).

Question 15

If $0<x<1$ , which value is largest?

$x^{1/2}$
$x^{1/3}$
$x^{-1/2}$ (correct answer)
$x^{0}$

Explanation: For 0 < x < 1, a negative exponent produces a reciprocal greater than 1, while every nonnegative power listed stays at or below 1. So x^{-1/2} is largest. The tempting choice x^{1/3} is still less than 1, so it cannot be the largest.

Question 16

Which of the following is equal to $\sqrt[4]{81}$ ?

$9$
$3$ (correct answer)
$\sqrt{81}$
$81^{1/2}$
$27$

Explanation: This question tests simplifying fourth roots using prime factorization. The relevant rule is that (sqrt[4]{81} =

81^{1/4}

), and (81 =

3^4

), so ((3^4)^{1/4} = 3). This is the principal real root. The result is justified as (

3^4

= 81), confirming the value. A distractor like choice A, 9, fails by confusing with the square root, as (

\sqrt{81}

= 9), not the fourth root.

Question 17

If $b$ is a real number and $b^{\frac{1}{3}}=-2$ , what is the value of $b$ ?

$-6$
$8$
$-8$ (correct answer)
$\pm 8$
$4$

Explanation: This question tests understanding of fractional exponents and cube roots. The rule states that

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

, which means we're looking for a number whose cube root is

-2

. If

b^{1/3} = -2

, then cubing both sides gives

(b^{1/3})^3 = (-2)^3

. By the power rule,

(b^{1/3})^3 = b^{3/3} = b^1 = b

. Therefore,

b = (-2)^3 = -8

. We can verify:

(-8)^{1/3} = \sqrt[3]{-8} = -2

✓ since the cube root of a negative number is negative. A common error would be to think cube roots of negative numbers don't exist, similar to square roots.

Question 18

If $x$ is a real number and $\sqrt[4]{x}=\dfrac{1}{3}$ , what is the value of $x^{3/4}$ ?

$\dfrac{1}{81}$
$\dfrac{1}{27}$ (correct answer)
$\dfrac{1}{9}$
$\dfrac{1}{3}$
$\dfrac{1}{243}$

Explanation: This question tests your understanding of fractional exponents and roots. Given

\sqrt[4]{x} = \frac{1}{3}

, which means

x^{1/4} = \frac{1}{3}

, we need to find

x^{3/4}

. First, we can find

x

by raising both sides to the fourth power:

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

, giving us

x = \frac{1}{81}

. Now we calculate

x^{3/4} = (\frac{1}{81})^{3/4} = \frac{1}{81^{3/4}} = \frac{1}{(81^{1/4})^3} = \frac{1}{3^3} = \frac{1}{27}

. Alternatively, since

x^{3/4} = (x^{1/4})^3 = (\frac{1}{3})^3 = \frac{1}{27}

. A common error would be to compute

x^{3/4}

\frac{3}{4} \times \frac{1}{81}

, which incorrectly multiplies instead of applying the exponent.

Question 19

If $y$ is a real number and $\sqrt{y+5}=4$ , what is the value of $y$ ?

$-1$
$11$ (correct answer)
$21$
$-21$
$\pm 11$

Explanation: This question tests solving equations involving square roots and understanding domain restrictions. The relevant rule is that (

\sqrt{y + 5}

= 4) implies (y + 5 geq 0) and the square root is non-negative, so squaring both sides gives (y + 5 = 16). Solving, (y = 11). This satisfies the original equation since (

\sqrt{11 + 5}

\sqrt{16}

= 4). The result is justified as the unique real solution meeting the non-negative root condition. A distractor like choice A, -1, might come from incorrectly solving (y + 5 = -4), but this ignores that square roots cannot equal negative values.

Question 20

If $m$ is a real number, $m\neq 0$ , and $m^{\frac{1}{2}}=3$ , what is the value of $m^{\frac{3}{2}}$ ?

$6$
$9$
$27$ (correct answer)
$81$
$\pm 27$

Explanation: This question tests the power rule for exponents when raising a power to another power. Given that

m^{1/2} = 3

, we need to find

m^{3/2}

. We can rewrite

m^{3/2}

(m^{1/2})^3

using the rule

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

. Since

m^{1/2} = 3

, we have

m^{3/2} = (m^{1/2})^3 = 3^3 = 27

. We can verify by first finding

m

: if

m^{1/2} = 3

, then

m = 9

, and

9^{3/2} = (9^{1/2})^3 = 3^3 = 27

✓. A common error would be to multiply the exponents incorrectly, thinking

m^{3/2} = m^{1/2} \cdot 3 = 3 \cdot 3 = 9