GRE Quantitative Quiz: Exponents Roots

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.

Question 1

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

Explanation: This question tests composing exponents from root equations. The relevant rule is that ($\sqrt{x}$ = $x^{1/2}$ = rac{1}{3}), so (x = left(rac{1}{3} ight)^2 = rac{1}{9}). Then, (x^{3/2} = (x^{1/2})^3 = left(rac{1}{3} ight)^3 = rac{1}{27}). Alternatively, (left(rac{1}{9} ight)^{3/2} = rac{1}{9} cdot sqrt{rac{1}{9}} = rac{1}{9} cdot rac{1}{3} = rac{1}{27}). The result is justified by consistent application of exponents. A distractor like choice A, (rac{1}{9}), fails by stopping at (x) instead of applying the 3/2 exponent.

Question 2

1. $\dfrac{1}{8}$ (correct answer)
2. $\dfrac{1}{4}$
3. $\dfrac{1}{2}$
4. $\dfrac{1}{16}$
5. $8$

Explanation: This question tests evaluating fractional exponents on fractions. The relevant rule is that (left(rac{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} = rac{1}{8}). Alternatively, (left(rac{1}{16} ight)^{1/4} = rac{1}{2}), then raised to the third power is (left(rac{1}{2} ight)^3 = rac{1}{8}). The result is justified by consistent exponent simplification. A distractor like choice B, (rac{1}{4}), fails by using an incorrect exponent like 1/2 instead of 3/4.

Question 3

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

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 (rac{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, (rac{1}{3}), fails by inverting the ratio, perhaps from miswriting (8 = $2^{1/3}$).

Question 4

1. $\sqrt{9}$
2. $\sqrt{16}$
3. $3$ (correct answer)
4. $9$
5. $6$

Explanation: This question tests simplifying ratios of square roots. The relevant rule is that (rac{$\sqrt{18}$}{$\sqrt{2}$} = \sqrt{rac{18}{2}} = $\sqrt{9}$). Simplifying further, ($\sqrt{9}$ = 3). Alternatively, factor as (rac{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 5

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

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 6

1. $2^2$ (correct answer)
2. $2^3$
3. $2^{5}$
4. $2^{2/2}$
5. $\sqrt{2^4}$

Explanation: This question tests subtracting exponents with the same base. The relevant rule is that (rac{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 7

1. $2x^2$
2. $4|x|$
3. $|4x^2|$
4. $4x^2$ (correct answer)
5. $4x$

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 $(x^4)^{1/2} = |x^2|$ or $2|x|$, not recognizing that $x^2$ is already non-negative.

Question 8

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

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 9

1. $\sqrt[3]{a^3}$
2. $a^2$
3. $a$ (correct answer)
4. $\sqrt[3]{a^2}$
5. $\sqrt[3]{a}$

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 10

1. $x$
2. $|x|$ (correct answer)
3. $\pm x$
4. $x^2$
5. $\dfrac{1}{|x|}$

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

1. $9$
2. $3$ (correct answer)
3. $\sqrt{81}$
4. $81^{1/2}$
5. $27$

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 12

1. $-6$
2. $8$
3. $-8$ (correct answer)
4. $\pm 8$
5. $4$

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 13

1. $\dfrac{1}{81}$
2. $\dfrac{1}{27}$ (correct answer)
3. $\dfrac{1}{9}$
4. $\dfrac{1}{3}$
5. $\dfrac{1}{243}$

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 $x^{3/4}$ as $\frac{3}{4} \times \frac{1}{81}$, which incorrectly multiplies instead of applying the exponent.

Question 14

1. $-1$
2. $11$ (correct answer)
3. $21$
4. $-21$
5. $\pm 11$

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 15

1. $6$
2. $9$
3. $27$ (correct answer)
4. $81$
5. $\pm 27$

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 $m^{3/2}$ as $(m^{1/2})^3$ 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 $m^{3/2} = m^{1/2} \cdot 3 = 3 \cdot 3 = 9$.

Question 16

1. $-4$
2. $4$ (correct answer)
3. $-8$
4. $8$
5. $\pm 4$

Explanation: This question tests composing fractional exponents from root equations. The relevant rule is that (sqrt[3]{r} = $r^{1/3}$ = -2), so (r = (-2)^3 = -8). Then, (r^{2/3} = (r^{1/3})^2 = (-2)^2 = 4). Alternatively, ((r^2)^{1/3} = (64)^{1/3} = 4). The result is justified by the even power making it positive. A distractor like choice A, -4, fails by not squaring the cube root, keeping the negative sign incorrectly.

Question 17

1. $\pm 4$
2. $-4$
3. $\sqrt[3]{8}$
4. $4$ (correct answer)
5. $2$

Explanation: This question tests the ability to solve equations involving fractional exponents. The fractional exponent rule states that $x^{m/n} = (x^{1/n})^m = \sqrt[n]{x^m}$, so $x^{3/2} = (x^{1/2})^3 = (\sqrt{x})^3$. Given that $x^{3/2} = 8$, we can rewrite this as $(\sqrt{x})^3 = 8$. Taking the cube root of both sides gives $\sqrt{x} = 2$, and squaring both sides yields $x = 4$. We can verify: $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$ ✓. A common error would be to think $x^{3/2} = 8$ means $x = 8^{2/3}$, which would give $x = 4$ anyway, but students might incorrectly compute $8^{2/3}$ as $\pm 4$.

Question 18

1. $b=9$
2. $b=-9$
3. There is no real number $b$ that satisfies the equation. (correct answer)
4. $b=3$
5. $b=\pm 9$

Explanation: This question tests the domain of square roots for real numbers. The relevant rule is that ($b^{1/2}$ = $\sqrt{b}$) is defined only for (b geq 0) and yields a non-negative result. For ($\sqrt{b}$ = -3), no real (b) works since the square root cannot be negative. Attempting to solve gives (b = (-3)^2 = 9), but ($\sqrt{9}$ = 3 eq -3). The result is justified by the fundamental property of square roots being non-negative. A distractor like choice A, (b=9), fails by ignoring that the square root of 9 is positive, not negative.

Question 19

1. $3$ (correct answer)
2. $-3$
3. $\pm 3$
4. $9$
5. $\pm 9$

Explanation: This question tests the principal square root and its relation to absolute value. The relevant rule is that ($\sqrt{a^2}$ = |a|), as the square root function returns the non-negative root. Given ($a^2$ = 9), (a = pm 3), so ($\sqrt{a^2}$ = | pm 3 | = 3). This applies regardless of the sign of (a). The result is justified because the square root of 9 is always 3, not considering signs. A distractor like choice C, (pm 3), fails by confusing the solutions to ($a^2$ = 9) with the non-negative value of the square root.

Question 20

1. $9$ (correct answer)
2. $-9$
3. $\pm 9$
4. $3$
5. $27$

Explanation: This question tests solving equations with fractional exponents involving even roots. The relevant rule is that (t^{3/2} = (t^{1/2})^3 = 27), where ($t^{1/2}$ geq 0). Let (s = $t^{1/2}$), so ($s^3$ = 27), (s = 3), then (t = 9). Negative (t) makes the square root undefined in reals. The result is justified as (9^{3/2} = (3^2)^{3/2} = 3^3 = 27). A distractor like choice B, -9, fails by not considering that even roots require non-negative bases in real numbers.