Exponents and Roots - Quantitative Reasoning

If $b$ is a real number and $b^{1/2}=-3$ , which of the following must be true?
Answer: There is no real number $b$ that satisfies the equation. 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.
If $a$ is a real number and $\sqrt{a^2}=7$ , which of the following must be true?
Answer: $a=\pm 7$ This question tests understanding of the principal square root and absolute value. The key rule is that $\sqrt{a^2} = |a|$ for any real number $a$ . Given that $\sqrt{a^2} = 7$ , we have $|a| = 7$ . By the definition of absolute value, this means $a = 7$ or $a = -7$ , which can be written as $a = \pm 7$ . Both values satisfy the original equation: $\sqrt{7^2} = \sqrt{49} = 7$ ✓ and $\sqrt{(-7)^2} = \sqrt{49} = 7$ ✓. A common error would be to think $\sqrt{a^2} = a$ , which would incorrectly lead to only $a = 7$ , missing the negative solution.
Which of the following is equal to $\sqrt[3]{a^4}$ for real $a$ ?
Answer: $a^{4/3}$ This question tests the equivalence of root expressions to exponential forms for real numbers. The relevant rule is that (sqrt[3] ${a^4$ } = $a^{4/3}$ ), as the cube root of a power directly translates to a fractional exponent. Applying this, (sqrt[3]{a^4} = $(a^4)^{1/3}$ = $a^{4/3}$ ). This holds for all real (a), including negative values, since cube roots are defined for negatives and yield the correct sign. The result is justified because both expressions produce the same value, such as for (a = -8), both give 16. A distractor like choice C, (sqrt[3]{a} sqrt[3]{a} = $a^{2/3}$ ), fails by using an incorrect exponent, resulting in ( $a^{2/3}$ ) instead of ( $a^{4/3}$ ).
If $a$ is a real number and $a^2=9$ , what is the value of $\sqrt{a^2}$ ?
Answer: $3$ 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.
Which of the following expressions is equivalent to $\sqrt[3]{a^2}\,\sqrt[3]{a}$ for real $a$ ?
Answer: $a$ 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.
Which of the following is equal to $\dfrac{\sqrt{18}}{\sqrt{2}}$ ?
Answer: $3$ 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.
If $n<0$ , which equals $\sqrt[3]{n^6}$ ?
Answer: $n^2$ 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).
If $x$ is a real number and $\sqrt{x}=\dfrac{1}{3}$ , what is the value of $x^{3/2}$ ?
Answer: $\dfrac{1}{27}$ 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}\right)^2 = \frac{1}{9}). Then, ( $x^{3/2}$ = $(x^{1/2})^3$ = left(\frac{1}{3}\right)^3 = \frac{1}{27}). Alternatively, (left(\frac{1}{9}\right)^{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.
Which of the following expressions is equivalent to $\left(16x^4\right)^{\frac{1}{2}}$ for real $x$ ?
Answer: $4x^2$ 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.
If $4^x=5$ , what is $2^{-2x}$ ?
Answer: 1/5 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.
If $y$ is a real number and $\sqrt{y+5}=4$ , what is the value of $y$ ?
Answer: $11$ 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.
Which of the following is equivalent to $\dfrac{\sqrt{x^6}}{x^2}$ for real $x\ne 0$ ?
Answer: $|x|$ 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.
If $x$ is a real number and $x^{1/3}=-2$ , what is the value of $x^{4/3}$ ?
Answer: $16$ This question tests your ability to work with fractional exponents when the base is negative. Given $x^{1/3} = -2$ , we need to find $x^{4/3}$ . First, we can find $x$ by cubing both sides: $(x^{1/3})^3 = (-2)^3$ , which gives us $x = -8$ . Now we can calculate $x^{4/3} = (-8)^{4/3} = ((-8)^{1/3})^4 = (-2)^4 = 16$ . Alternatively, we can use the fact that $x^{4/3} = x^{1/3} \cdot x^{3/3} = x^{1/3} \cdot x = (-2) \cdot (-8) = 16$ . A common error would be to think the result should be negative, but raising a negative number to an even power (4 in this case) always gives a positive result.
Which of the following is equivalent to $\dfrac{2^{5/2}}{2^{1/2}}$ ?
Answer: $2^2$ 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.
If $m$ is a real number, $m\neq 0$ , and $m^{\frac{1}{2}}=3$ , what is the value of $m^{\frac{3}{2}}$ ?
Answer: $27$ 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 $m^{3/2} = m^{1/2} \cdot 3 = 3 \cdot 3 = 9$ .
Which of the following is equivalent to $\left(\dfrac{1}{16}\right)^{3/4}$ ?
Answer: $\dfrac{1}{8}$ This question tests evaluating fractional exponents on fractions. The relevant rule is that (left(\frac{1}{16}\right)^{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}\right)^{1/4} = \frac{1}{2}), then raised to the third power is (left(\frac{1}{2}\right)^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.
If $b$ is a real number and $b^{\frac{1}{3}}=-2$ , what is the value of $b$ ?
Answer: $-8$ 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.
If $x^{\frac{3}{2}}=8$ and $x$ is a real number, what is the value of $x$ ?
Answer: $4$ 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 $\pm 4$ .
Which of the following is equal to $\left(\sqrt[3]{5}\right)^2$ ?
Answer: $5^{2/3}$ This question tests rewriting root expressions in exponential form. The relevant rule is that (left(sqrt[3]{5}\right)^2 = $(5^{1/3})^2$ = $5^{2/3}$ ). This combines the exponents correctly. The expression cannot be simplified further without a calculator. The result is justified as it matches the power rule for exponents. A distractor like choice D, ( $5^{3/2}$ ), fails by swapping the numerator and denominator of the exponent, resulting in a different value.
Which of the following is true for all real numbers $p$ such that $p\ne 0$ ?
Answer: $(p^3)^{1/3}=p$ This question tests properties of exponents and roots that hold for all real non-zero numbers. The relevant rule is that ( $(p^3)^{1/3}$ = $p^{3 cdot (1/3)}$ = p), and cube roots are defined for all reals. This equality holds with correct signs for negative (p). For example, if (p = -8), ( $(-512)^{1/3}$ = -8 = p). The result is justified as the exponents multiply correctly without domain issues. A distractor like choice B, ( $(p^2)^{1/2}$ = |p|), fails by equaling |p| instead of p for negative values.

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