Radical & Rational Exponent Forms - Math 2

If $\sqrt{\sqrt[3]{a}} = a^k$ , then what is the value of $k$ ?
Answer: $\frac{1}{6}$ When you encounter nested radicals and exponents like this, you need to convert everything to exponential form and use the laws of exponents to solve for the unknown. Start by rewriting the radicals as fractional exponents. The cube root $\sqrt[3]{a}$ becomes $a^{1/3}$ , and the square root of that becomes $(a^{1/3})^{1/2}$ . Using the power rule $(a^m)^n = a^{mn}$ , this simplifies to $a^{(1/3)(1/2)} = a^{1/6}$ . Since $\sqrt{\sqrt[3]{a}} = a^k$ , we have $a^{1/6} = a^k$ . For this equation to hold for all valid values of $a$ , the exponents must be equal, so $k = \frac{1}{6}$ . This makes D the correct answer. Looking at the wrong answers: A) $\frac{3}{2}$ likely comes from incorrectly multiplying the denominators (3 × 2 = 6, then flipping to get 3/2). B) $\frac{2}{3}$ might result from confusing which root goes where or incorrectly applying the multiplication rule. C) $\frac{5}{6}$ could come from adding the fractions $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ instead of multiplying them. Remember this key strategy: when dealing with nested radicals, always convert to fractional exponents first, then apply the power rule by multiplying exponents. The pattern is that $\sqrt[n]{\sqrt[m]{a}} = a^{1/(mn)}$ , not $a^{1/(m+n)}$ .
If $\sqrt[n]{x^m} = x^{\frac{2}{5}}$ for all positive values of $x$ , and $m = 6$ , what is the value of $n$ ?
Answer: 15 Since $\sqrt[n]{x^m} = x^{\frac{m}{n}}$ , we have $x^{\frac{m}{n}} = x^{\frac{2}{5}}$ . This means $\frac{m}{n} = \frac{2}{5}$ . Given that $m = 6$ , we substitute: $\frac{6}{n} = \frac{2}{5}$ . Cross-multiplying: $6 \cdot 5 = 2 \cdot n$ , so $30 = 2n$ , which gives $n = 15$ . Choice A comes from incorrectly calculating $6 \cdot 2 = 12$ . Choice C comes from incorrectly calculating $6 + 12 = 18$ . Choice D comes from incorrectly calculating $5 \cdot 4 = 20$ .
Which of the following expressions is equivalent to $\sqrt[6]{x^3} \cdot \sqrt[3]{x}$ in rational exponent form?
Answer: $x^{\frac{5}{6}}$ Convert each radical to rational exponent form: $\sqrt[6]{x^3} = x^{\frac{3}{6}} = x^{\frac{1}{2}}$ and $\sqrt[3]{x} = x^{\frac{1}{3}}$ . Then multiply using the rule $x^a \cdot x^b = x^{a+b}$ : $x^{\frac{1}{2}} \cdot x^{\frac{1}{3}} = x^{\frac{1}{2} + \frac{1}{3}}$ . Finding a common denominator: $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$ . Choice B comes from incorrectly adding $\frac{3}{6} + \frac{4}{6}$ . Choice C comes from multiplying the exponents instead of adding them. Choice D comes from incorrectly converting $\sqrt[3]{x}$ to $x^3$ instead of $x^{\frac{1}{3}}$ .
If $x^{\frac{2}{3}} \cdot x^{\frac{1}{4}} = \sqrt[12]{x^k}$ for all positive values of $x$ , what is the value of $k$ ?
Answer: 11 First, simplify the left side using the rule $x^a \cdot x^b = x^{a+b}$ : $x^{\frac{2}{3}} \cdot x^{\frac{1}{4}} = x^{\frac{2}{3} + \frac{1}{4}} = x^{\frac{8}{12} + \frac{3}{12}} = x^{\frac{11}{12}}$ . Next, convert the right side to rational exponent form: $\sqrt[12]{x^k} = x^{\frac{k}{12}}$ . Setting the exponents equal: $\frac{11}{12} = \frac{k}{12}$ , so $k = 11$ . Choice A (7) comes from incorrectly subtracting exponents. Choice B (8) comes from finding the LCD but forgetting to add the numerators. Choice D (12) comes from confusing the root index with the answer.
If $y = \sqrt[5]{x^3}$ and $z = x^{\frac{2}{5}}$ , then $\frac{y^2}{z}$ can be written as $x^n$ . What is the value of $n$ ?
Answer: $\frac{4}{5}$ This question tests your ability to manipulate expressions with rational exponents. When you see fractional exponents and radicals together, the key is to convert everything to exponential form using the rules $\sqrt[n]{x} = x^{1/n}$ and $(x^a)^b = x^{ab}$ . Let's rewrite both expressions using rational exponents. Since $y = \sqrt[5]{x^3}$ , we can write this as $y = (x^3)^{1/5} = x^{3/5}$ . The expression $z = x^{2/5}$ is already in exponential form. Now we can find $\frac{y^2}{z}$ . First, $y^2 = (x^{3/5})^2 = x^{(3/5) \cdot 2} = x^{6/5}$ . Then: $\frac{y^2}{z} = \frac{x^{6/5}}{x^{2/5}} = x^{6/5 - 2/5} = x^{4/5}$ So $n = \frac{4}{5}$ , which is answer choice C. Looking at the wrong answers: Choice A ( $\frac{1}{25}$ ) likely comes from incorrectly multiplying the denominators 5 and 5 to get 25. Choice B ( $\frac{6}{25}$ ) might result from getting the correct numerator 6 but the wrong denominator 25. Choice D ( $\frac{11}{25}$ ) could come from adding exponents incorrectly, perhaps computing $\frac{6}{5} + \frac{2}{5} = \frac{8}{5}$ instead of subtracting, then making the denominator error. Remember: when dividing powers with the same base, subtract the exponents. Always convert radicals to exponential form first to avoid computational errors.
If $8^{\frac{2}{3}} \cdot 2^x = \sqrt[3]{32}$ , what is the value of $x$ ?
Answer: $-\frac{1}{3}$ First, convert everything to the same base. Since $8 = 2^3$ and $32 = 2^5$ , we have: $8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4$ and $\sqrt[3]{32} = 32^{\frac{1}{3}} = (2^5)^{\frac{1}{3}} = 2^{\frac{5}{3}}$ . The equation becomes: $4 \cdot 2^x = 2^{\frac{5}{3}}$ , or $2^2 \cdot 2^x = 2^{\frac{5}{3}}$ , which gives $2^{2+x} = 2^{\frac{5}{3}}$ . Therefore: $2 + x = \frac{5}{3}$ , so $x = \frac{5}{3} - 2 = \frac{5}{3} - \frac{6}{3} = -\frac{1}{3}$ . Choice B comes from forgetting the negative sign. Choice C comes from incorrectly calculating $\frac{5}{3} - 1$ instead of $\frac{5}{3} - 2$ . Choice D comes from setting $x = \frac{5}{3} - \frac{2}{3}$ .
If $x^{\frac{3}{4}} = 8$ , then what is the value of $x^{\frac{9}{8}}$ ?
Answer: $16\sqrt{2}$ First, solve for x from $x^{\frac{3}{4}} = 8$ . Raising both sides to the power $\frac{4}{3}$ : $x = 8^{\frac{4}{3}} = (2^3)^{\frac{4}{3}} = 2^4 = 16$ . Then $x^{\frac{9}{8}} = 16^{\frac{9}{8}} = (2^4)^{\frac{9}{8}} = 2^{\frac{9}{2}} = 2^4 \cdot 2^{\frac{1}{2}} = 16\sqrt{2}$ . Choice B results from incorrectly computing $2^{\frac{9}{2}}$ as $32\sqrt{2}$ . Choices C and D result from computational errors in the exponent manipulation.
If $\sqrt[4]{16x^8y^{12}} = 2x^ay^b$ for all non-negative values of $x$ and $y$ , what is the value of $a + b$ ?
Answer: 5 Convert the radical to rational exponent form: $\sqrt[4]{16x^8y^{12}} = (16x^8y^{12})^{\frac{1}{4}} = 16^{\frac{1}{4}} \cdot x^{8 \cdot \frac{1}{4}} \cdot y^{12 \cdot \frac{1}{4}} = 2x^2y^3$ . Therefore, $a = 2$ and $b = 3$ , so $a + b = 5$ . Choice A comes from incorrectly calculating $b = 1$ instead of $b = 3$ . Choice B comes from miscalculating $16^{\frac{1}{4}} = 4$ instead of 2. Choice D comes from adding the original exponents $8 + 12 = 20$ without applying the fourth root operation.
Which expression is equivalent to $\frac{\sqrt[3]{x^5}}{\sqrt{x^3}}$ for $x > 0$ ?
Answer: $x^{\frac{1}{6}}$ Convert to rational exponents: $\frac{\sqrt[3]{x^5}}{\sqrt{x^3}} = \frac{x^{\frac{5}{3}}}{x^{\frac{3}{2}}} = x^{\frac{5}{3} - \frac{3}{2}} = x^{\frac{10}{6} - \frac{9}{6}} = x^{\frac{1}{6}}$ . Choice B results from adding exponents instead of subtracting. Choice C comes from incorrectly computing $\frac{5}{3} - \frac{3}{2}$ as $\frac{5}{6}$ . Choice D results from adding the original exponents: $\frac{5}{3} + \frac{3}{2} = \frac{11}{6}$ .
Which expression represents $\left(\frac{1}{x^{\frac{3}{4}}}\right)^{\frac{4}{3}}$ in simplest radical form?
Answer: $\frac{1}{x}$ First, rewrite using negative exponents: $\left(\frac{1}{x^{\frac{3}{4}}}\right)^{\frac{4}{3}} = \left(x^{-\frac{3}{4}}\right)^{\frac{4}{3}}$ . Using the power rule $(x^a)^b = x^{ab}$ : $x^{-\frac{3}{4} \cdot \frac{4}{3}} = x^{-\frac{12}{12}} = x^{-1} = \frac{1}{x}$ . Choice A comes from incorrectly calculating the exponent as $-\frac{1}{2}$ . Choice C comes from incorrectly calculating the exponent as $-\frac{1}{3}$ . Choice D comes from incorrectly calculating the exponent as $-2$ .
If $27^{\frac{x}{3}} = \sqrt[4]{9}$ , what is the value of $x$ ?
Answer: $\frac{1}{2}$ This problem tests your ability to work with exponential equations involving fractional exponents and radicals. When you encounter equations like this, the key strategy is to express both sides using the same base, then equate the exponents. Start by rewriting both sides in terms of base 3. Since $27 = 3^3$ , the left side becomes $(3^3)^{\frac{x}{3}} = 3^{3 \cdot \frac{x}{3}} = 3^x$ . For the right side, note that $9 = 3^2$ , so $\sqrt[4]{9} = 9^{\frac{1}{4}} = (3^2)^{\frac{1}{4}} = 3^{2 \cdot \frac{1}{4}} = 3^{\frac{1}{2}}$ . Now your equation becomes $3^x = 3^{\frac{1}{2}}$ . Since the bases are equal, the exponents must be equal: $x = \frac{1}{2}$ . This confirms answer D. Let's check why the other options are incorrect. Answer A ( $x = 3$ ) would give $27^1 = 27$ on the left side, but $\sqrt[4]{9} \approx 1.73$ , so these aren't equal. Answer B ( $x = \frac{3}{2}$ ) yields $27^{\frac{1}{2}} = \sqrt{27} \approx 5.2$ , which also doesn't match $\sqrt[4]{9}$ . Answer C ( $x = 2$ ) produces $27^{\frac{2}{3}} = (27^{\frac{1}{3}})^2 = 3^2 = 9$ , again not equal to $\sqrt[4]{9}$ . When solving exponential equations, always aim to express both sides with the same base. This transforms the problem into a simple equation involving exponents, making it much easier to solve.
Which expression is equivalent to $\frac{1}{\sqrt[3]{x^2}}$ for $x \neq 0$ ?
Answer: $x^{-\frac{2}{3}}$ When you encounter radical expressions with variables, the key is converting between radical notation and exponential notation using the fundamental relationship: $\sqrt[n]{x^m} = x^{\frac{m}{n}}$ and $\frac{1}{x^a} = x^{-a}$ . Let's work through this step by step. The expression $\frac{1}{\sqrt[3]{x^2}}$ has a cube root in the denominator. First, convert the cube root to exponential form: $\sqrt[3]{x^2} = x^{\frac{2}{3}}$ . Now your expression becomes $\frac{1}{x^{\frac{2}{3}}}$ . When you have a fraction with 1 in the numerator and a power in the denominator, you flip the sign of the exponent: $\frac{1}{x^{\frac{2}{3}}} = x^{-\frac{2}{3}}$ . This confirms that choice D is correct. Looking at the wrong answers: Choice A gives $x^{\frac{2}{3}}$ , which is what you'd get if you forgot to account for the fraction bar—this is just the cube root without the reciprocal. Choice B gives $x^{\frac{3}{2}}$ , which flips the numerator and denominator in the exponent, a common error when converting radicals. Choice C gives $x^{-\frac{3}{2}}$ , which combines the reciprocal error from A with the flipped fraction error from B. Remember this pattern: when converting $\frac{1}{\sqrt[n]{x^m}}$ , first write the radical as $x^{\frac{m}{n}}$ , then apply the negative exponent rule to get $x^{-\frac{m}{n}}$ . The fraction stays in the same order as the original radical.
Which of the following expressions is equivalent to $(8x^6)^{\frac{2}{3}}$ when written in radical form?
Answer: $4x^4$ When you encounter expressions with fractional exponents, remember that these can be converted between exponential and radical forms using the relationship $a^{\frac{m}{n}} = \sqrt[n]{a^m}$ . To solve $(8x^6)^{\frac{2}{3}}$ , first apply the power rule: raise each factor inside the parentheses to the $\frac{2}{3}$ power. This gives us $8^{\frac{2}{3}} \cdot (x^6)^{\frac{2}{3}}$ . For $8^{\frac{2}{3}}$ : Since $8 = 2^3$ , we have $(2^3)^{\frac{2}{3}} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4$ . For $(x^6)^{\frac{2}{3}}$ : Using the power rule again, $x^{6 \cdot \frac{2}{3}} = x^4$ . Therefore, $(8x^6)^{\frac{2}{3}} = 4x^4$ . Looking at the wrong answers: Choice A, $\sqrt[3]{64x^{12}}$ , equals $4x^4$ when simplified, but the question asks for radical form and this isn't the most direct conversion. Choice B, $2\sqrt[3]{8x^6}$ , incorrectly applies the fractional exponent—this would equal $2 \cdot 2x^2 = 4x^2$ , not $4x^4$ . Choice D, $4x^4\sqrt[3]{1}$ , is mathematically equivalent to $4x^4$ since $\sqrt[3]{1} = 1$ , but it's unnecessarily complex. The answer is C: $4x^4$ . Strategy tip: When working with fractional exponents, always check if the base is a perfect power (like $8 = 2^3$ ). This often allows for clean simplification without needing radical notation.
Which of the following is equivalent to $\sqrt[6]{x^4y^9}$ written using rational exponents?
Answer: $x^{\frac{2}{3}}y^{\frac{3}{2}}$ $\sqrt[6]{x^4y^9} = (x^4y^9)^{\frac{1}{6}} = x^{4 \cdot \frac{1}{6}} \cdot y^{9 \cdot \frac{1}{6}} = x^{\frac{4}{6}} \cdot y^{\frac{9}{6}} = x^{\frac{2}{3}} \cdot y^{\frac{3}{2}}$ . Choice A is correct but not fully simplified using rational exponent properties. Choice C is correct but not simplified ( $\frac{4}{6} = \frac{2}{3}$ and $\frac{9}{6} = \frac{3}{2}$ ). Choice D incorrectly applies the radical to each variable separately with the same exponent.
Which of the following is equivalent to $\frac{\sqrt[4]{x^3}}{\sqrt[3]{x^2}}$ when $x > 0$ ?
Answer: $x^{\frac{1}{12}}$ Convert to rational exponents: $\frac{\sqrt[4]{x^3}}{\sqrt[3]{x^2}} = \frac{x^{\frac{3}{4}}}{x^{\frac{2}{3}}}$ . Using the quotient rule $\frac{x^a}{x^b} = x^{a-b}$ : $x^{\frac{3}{4} - \frac{2}{3}}$ . Finding a common denominator: $\frac{3}{4} - \frac{2}{3} = \frac{9}{12} - \frac{8}{12} = \frac{1}{12}$ . Choice B comes from incorrectly calculating $\frac{9}{12} - \frac{4}{12} = \frac{5}{12}$ . Choice C comes from adding instead of subtracting: $\frac{3}{4} + \frac{2}{3} = \frac{9}{12} + \frac{8}{12} = \frac{17}{12}$ , then making an arithmetic error. Choice D comes from incorrectly calculating $\frac{9}{12} + \frac{2}{12} = \frac{11}{12}$ .
Which expression is equivalent to $\frac{1}{\sqrt[3]{a^2b}}$ when $a > 0$ and $b > 0$ ?
Answer: $(a^2b)^{-\frac{1}{3}}$ First, rewrite the radical in rational exponent form: $\sqrt[3]{a^2b} = (a^2b)^{\frac{1}{3}}$ . Then, since we have $\frac{1}{\sqrt[3]{a^2b}}$ , this becomes $\frac{1}{(a^2b)^{\frac{1}{3}}} = (a^2b)^{-\frac{1}{3}}$ . Choice A incorrectly distributes the exponent 2 to both $a$ and $b$ . Choice C correctly applies the negative exponent rule but incorrectly separates $a^2$ and $b$ before applying the $\frac{1}{3}$ exponent. Choice D forgets the negative exponent entirely.
If $\left(\sqrt[5]{a^2}\right)^{10} = a^n$ , what is the value of $n$ ?
Answer: $4$ This question tests your ability to work with exponents and radicals, specifically how to convert between radical notation and fractional exponents. When you see nested exponents and radicals, the key is to systematically convert everything to exponential form and apply the power rules. First, convert the radical to exponential form: $\sqrt[5]{a^2} = a^{2/5}$ . Now substitute this into the original expression: $\left(a^{2/5}\right)^{10} = a^n$ . Using the power rule $(a^m)^p = a^{mp}$ , you multiply the exponents: $\left(a^{2/5}\right)^{10} = a^{(2/5) \cdot 10} = a^{20/5} = a^4$ . Therefore, $n = 4$ . Looking at the wrong answers: Choice A ( $n = 2$ ) likely comes from only considering the exponent inside the radical and ignoring the outer exponent of 10. Choice B ( $n = 10$ ) represents taking just the outer exponent while ignoring the $\frac{2}{5}$ from the radical conversion. Choice C ( $n = 8$ ) might result from incorrectly adding the exponents ( $2 + 10 - 4 = 8$ ) rather than multiplying them according to the power rule. Remember that when working with nested exponents and radicals, always convert radicals to fractional exponents first ( $\sqrt[n]{a^m} = a^{m/n}$ ), then apply the power rule by multiplying exponents. This systematic approach prevents the common mistake of adding instead of multiplying exponents.
If $\sqrt[4]{a^3} \cdot \sqrt[6]{a^2} = a^k$ , then what is the value of $k$ ?
Answer: $\frac{13}{12}$ Convert to rational exponents: $\sqrt[4]{a^3} = a^{\frac{3}{4}}$ and $\sqrt[6]{a^2} = a^{\frac{2}{6}} = a^{\frac{1}{3}}$ . When multiplying powers with the same base, add exponents: $a^{\frac{3}{4}} \cdot a^{\frac{1}{3}} = a^{\frac{3}{4} + \frac{1}{3}} = a^{\frac{9}{12} + \frac{4}{12}} = a^{\frac{13}{12}}$ . Choice B results from subtracting instead of adding: $\frac{3}{4} - \frac{1}{3} = \frac{5}{12}$ (not $\frac{5}{6}$ ). Choice C comes from incorrect common denominator calculation. Choice D results from computing $\frac{3}{4} + \frac{2}{6}$ without simplifying $\frac{2}{6}$ first.
If $27^{\frac{2}{3}} = 3^m$ , what is the value of $m$ ?
Answer: $\frac{6}{3}$ First, rewrite 27 as a power of 3: $27 = 3^3$ . Then $27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{3 \cdot \frac{2}{3}} = 3^2$ . Therefore $m = 2 = \frac{6}{3}$ . Choice A results from incorrectly multiplying $27 \times \frac{2}{3} = 18$ then dividing by 3. Choice C comes from $27 \times 2 = 54$ divided by 9 (confusing the base). Choice D results from adding the exponents instead of multiplying: $3 + \frac{2}{3} = \frac{11}{3}$ (close to $\frac{8}{3}$ ).

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