Exponents & Roots - GMAT Quantitative Reasoning

For non-zero $a$ , the expression $\dfrac{a^{3}\,a^{-5}}{a^{-4}}$ can be written in the form $a^{k}$ . What is the value of $k$ ?
Answer: 2 When you encounter expressions with exponents that need to be simplified, you're working with the fundamental rules of exponents. The key insight is that you can combine terms with the same base by adding and subtracting exponents according to specific rules. Let's simplify $\dfrac{a^{3}\,a^{-5}}{a^{-4}}$ step by step. First, handle the numerator using the rule that when multiplying terms with the same base, you add the exponents: $a^{3} \cdot a^{-5} = a^{3+(-5)} = a^{-2}$ . Now you have $\dfrac{a^{-2}}{a^{-4}}$ . When dividing terms with the same base, you subtract the exponent in the denominator from the exponent in the numerator: $a^{-2} \div a^{-4} = a^{-2-(-4)} = a^{-2+4} = a^{2}$ . Therefore, $k = 2$ , making C correct. Looking at the wrong answers: A) -6 likely comes from incorrectly adding all the exponents: $3 + (-5) + (-4) = -6$ , but this ignores the division operation. B) -2 represents stopping after combining the numerator ( $a^{3} \cdot a^{-5} = a^{-2}$ ) without completing the division. D) 12 might result from multiplying the exponents instead of adding/subtracting them, which violates the basic rules of exponents. Study tip: Remember the exponent rules as "MADSPM" - when you Multiply, Add exponents; when you Divide, Subtract; when you raise a Power to a power, Multiply. Always work systematically from left to right, and double-check by substituting a simple value like $a = 2$ .
If $\sqrt[3]{2^x \cdot 4^y \cdot 8^z} = 16$ , what is the value of $x + 2y + 3z$ ?
Answer: $12$ When you encounter equations with exponents and radicals, your first step should be to express everything using the same base. This allows you to work with the exponents directly. Start by rewriting all terms using base 2: $4 = 2^2$ and $8 = 2^3$ . So the equation becomes: $\sqrt[3]{2^x \cdot (2^2)^y \cdot (2^3)^z} = 16$ Using the power rule $(a^m)^n = a^{mn}$ , this simplifies to: $\sqrt[3]{2^x \cdot 2^{2y} \cdot 2^{3z}} = 16$ When multiplying powers with the same base, add the exponents: $\sqrt[3]{2^{x+2y+3z}} = 16$ A cube root can be written as a fractional exponent: $2^{(x+2y+3z)/3} = 16$ Since $16 = 2^4$ , you have: $2^{(x+2y+3z)/3} = 2^4$ When the bases are equal, the exponents must be equal: $\frac{x+2y+3z}{3} = 4$ Therefore: $x+2y+3z = 12$ The answer is C) $12$ . Let's examine why the other choices are wrong. Choice A) $24$ would result if you incorrectly multiplied by 3 instead of dividing when solving the final equation. Choice B) $16$ represents the value on the right side of the original equation, a common trap for students who confuse the given value with the answer. Choice D) $36$ might result from calculation errors in the exponent manipulation. Strategy tip: Always convert to a common base in exponential equations, then use the property that equal bases mean equal exponents.
Which of the following equals $\sqrt[4]{32}$ ?
Answer: $2\sqrt[4]{2}$ When you encounter radical expressions on the GMAT, the key is to break down the radicand (the number under the radical) into its prime factors, then apply the radical properties systematically. To find $\sqrt[4]{32}$ , start by expressing 32 as a power of 2: $32 = 2^5$ . So $\sqrt[4]{32} = \sqrt[4]{2^5} = (2^5)^{1/4} = 2^{5/4}$ . Now rewrite this exponent: $2^{5/4} = 2^{1 + 1/4} = 2^1 \cdot 2^{1/4} = 2 \cdot 2^{1/4}$ . Since $2^{1/4} = \sqrt[4]{2}$ , we get $2\sqrt[4]{2}$ , which is choice B. Choice A gives $2^{1.2} = 2^{6/5}$ , which doesn't equal $2^{5/4}$ since $\frac{6}{5} = \frac{24}{20}$ while $\frac{5}{4} = \frac{25}{20}$ . Choice C gives $4\sqrt[4]{2} = 2^2 \cdot 2^{1/4} = 2^{2 + 1/4} = 2^{9/4}$ , which is too large. Choice D gives $\sqrt{8} = \sqrt{2^3} = 2^{3/2}$ . Converting to compare: $\frac{3}{2} = \frac{6}{4}$ while our answer is $2^{5/4}$ , so this is also incorrect. Strategy tip: When simplifying radicals, always convert to exponential form first ( $\sqrt[n]{a} = a^{1/n}$ ), use prime factorization, then convert back to radical form if needed. This systematic approach prevents calculation errors and makes comparisons between answer choices much clearer.
For positive $x$ , $\left(\dfrac{1}{x^{3}}\right)^{\,1/2}=x^{k}$ . What is $k$ ?
Answer: -\dfrac{3}{2} This question tests your understanding of exponent rules, particularly how to work with fractional exponents and negative powers. When you see nested exponents like this, the key is to systematically apply the power rules to simplify the expression. Starting with $\left(\dfrac{1}{x^{3}}\right)^{\,1/2}$ , first rewrite the fraction using negative exponents: $\dfrac{1}{x^{3}} = x^{-3}$ . This gives us $(x^{-3})^{1/2}$ . Now apply the power rule: when raising a power to another power, multiply the exponents. So $(x^{-3})^{1/2} = x^{(-3) \cdot (1/2)} = x^{-3/2}$ . Since this equals $x^k$ , we have $k = -\dfrac{3}{2}$ . Looking at the wrong answers: Choice (A) gives $k = -3$ , which would result from forgetting to apply the outer exponent of $\frac{1}{2}$ . Choice (C) gives $k = \frac{3}{2}$ , which reflects correctly applying the exponent rules but forgetting that $\frac{1}{x^3} = x^{-3}$ , not $x^3$ . Choice (D) gives $k = \frac{2}{3}$ , which comes from incorrectly multiplying the exponents as $\frac{1}{2} \cdot 3 = \frac{3}{2}$ but then inverting it, while also missing the negative sign. The correct answer is (B) $k = -\dfrac{3}{2}$ . Strategy tip: When working with nested exponents, always convert fractions to negative exponents first, then systematically apply the power rule by multiplying exponents. Double-check your signs—negative exponents from fractions often carry through the entire calculation.
For positive real numbers $k$ and $n$ satisfying $2^{k}=10^{n}$ , which of the following expresses $n$ in terms of $k$ ?
Answer: $k\log_{10}2$ When you encounter equations with different bases like $2^k = 10^n$ , you need to use logarithms to isolate the variables in the exponents. The key insight is choosing the right logarithm base to simplify your work. Starting with $2^k = 10^n$ , take the logarithm base 10 of both sides: $\log_{10}(2^k) = \log_{10}(10^n)$ Using the logarithm power rule, $\log_b(x^a) = a\log_b(x)$ : $k \cdot \log_{10}(2) = n \cdot \log_{10}(10)$ Since $\log_{10}(10) = 1$ : $k \cdot \log_{10}(2) = n$ Therefore, $n = k\log_{10}2$ , which is choice (A). Let's examine why the other options are incorrect: Choice (B) gives $k\log_2 10$ . This would be correct if we had solved for $k$ in terms of $n$ by taking $\log_2$ of both sides, but we need $n$ in terms of $k$ . Choice (C) gives $\frac{k}{\log_2 10}$ . This represents a common algebraic error—incorrectly manipulating the relationship between $k\log_2 10$ and the desired form. Choice (D) gives $\frac{k}{\log_{10} 2}$ . This might seem appealing because it contains $\log_{10} 2$ , but it incorrectly places this term in the denominator rather than as a coefficient. Strategy tip: When solving exponential equations with different bases, always take the logarithm using the base that appears in your desired answer format. Here, since we want $n$ (related to base 10), using $\log_{10}$ makes the algebra cleanest.
If $\sqrt[3]{8x^{6}}=2x^{m}$ for all positive real $x$ , what is the value of $m$ ?
Answer: 2 This question tests your understanding of exponent rules and radical simplification. When you see radicals that can be expressed as exponential expressions, your goal is to manipulate both sides to find unknown values. Start by simplifying the left side of the equation $\sqrt[3]{8x^{6}}=2x^{m}$ . The cube root can be written as a fractional exponent: $\sqrt[3]{8x^{6}} = (8x^{6})^{1/3}$ . Breaking this down further: $(8x^{6})^{1/3} = 8^{1/3} \cdot (x^{6})^{1/3}$ . Since $8 = 2^3$ , we have $8^{1/3} = (2^3)^{1/3} = 2^{3 \cdot 1/3} = 2^1 = 2$ . For the variable part, $(x^{6})^{1/3} = x^{6 \cdot 1/3} = x^2$ . Therefore, $\sqrt[3]{8x^{6}} = 2x^2$ , which means $2x^2 = 2x^{m}$ . Since the coefficients are equal and this must hold for all positive real $x$ , we need $x^2 = x^{m}$ , so $m = 2$ . Choice (A) 1/3 might tempt you if you confused the cube root's fractional exponent with the final answer. Choice (B) 3 could result from mistakenly thinking the cube root operation itself determines $m$ . Choice (D) 6 might appeal if you forgot to apply the fractional exponent rule and just used the original exponent from $x^6$ . When working with radicals and exponents, always convert radicals to fractional exponents first, then apply the power rule $(a^m)^n = a^{mn}$ systematically to avoid calculation errors.
If $\sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} = 5$ for $x > 0$ , what is the value of $x$ ?
Answer: $x = 20$ Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}$ . Since this is an infinite nested radical that converges, we have $y = \sqrt{x + y}$ . Given that $y = 5$ , we get $5 = \sqrt{x + 5}$ . Squaring both sides: $25 = x + 5$ , so $x = 20$ . We can verify: if $x = 20$ , then $y = \sqrt{20 + y}$ , and with $y = 5$ , we get $5 = \sqrt{20 + 5} = \sqrt{25} = 5$ ✓. Choice B comes from incorrectly setting up $y^2 = x$ instead of $y^2 = x + y$ . Choice C results from $5^2 + 5 = 30$ (adding instead of subtracting). Choice D comes from $5 \times 7 = 35$ (using an incorrect relationship).
Positive numbers $r, s, t$ satisfy $r^{2}=s^{3}=t^{5}=k$ . Which of the following is equal to $k^{\,1/30}$ ?
Answer: $r^{1/15}$ When you encounter equations with different exponents set equal to the same value, you're dealing with a problem that requires expressing everything in terms of a common base or finding equivalent fractional exponents. Given that $r^{2}=s^{3}=t^{5}=k$ , you can express each variable in terms of $k$ . From $r^{2}=k$ , you get $r=k^{1/2}$ . From $s^{3}=k$ , you get $s=k^{1/3}$ . From $t^{5}=k$ , you get $t=k^{1/5}$ . To find $k^{1/30}$ , you need to determine which answer choice equals this expression. The key insight is finding a common denominator for the exponents. Notice that 30 is the least common multiple of 2, 3, and 5. Let's check each option: Choice A: $r^{1/15} = (k^{1/2})^{1/15} = k^{1/30}$ . This matches exactly what we're looking for. Choice B: $s^{1/12} = (k^{1/3})^{1/12} = k^{1/36}$ . This gives the wrong exponent. Choice C: $t^{1/3} = (k^{1/5})^{1/3} = k^{1/15}$ . This exponent is too large. Choice D: $r^{1/10} = (k^{1/2})^{1/10} = k^{1/20}$ . This exponent is also too large. Only choice A produces the required exponent of $1/30$ . Study tip: When variables with different exponents equal the same value, express each variable as a power of that common value, then use exponent rules to manipulate the expressions. Always look for the least common multiple of the given exponents—it often appears in the target expression.
If $\sqrt{x^4 + 8x^2 + 16} = x^2 + 4$ for all real values of $x$ , which of the following must be true?
Answer: The equation is satisfied by all real numbers Notice that $x^4 + 8x^2 + 16 = (x^2 + 4)^2$ . Therefore, $\sqrt{x^4 + 8x^2 + 16} = \sqrt{(x^2 + 4)^2} = |x^2 + 4|$ . Since $x^2 \geq 0$ for all real $x$ , we have $x^2 + 4 \geq 4 > 0$ , so $|x^2 + 4| = x^2 + 4$ . Thus the equation becomes $x^2 + 4 = x^2 + 4$ , which is true for all real $x$ . Choice A is incorrect because $x$ can be negative. Choice B is true but not the best answer since it's always true regardless of the equation. Choice C is incorrect because all real numbers satisfy the equation.
If $\left(9^{x}\right)^{1/2}\,\times\,\left(27^{x}\right)^{1/3}=3^{10}$ , what is the value of $x$ ?
Answer: 5 When you encounter exponential equations with different bases, your first step should be to express everything using the same base. This allows you to work with the exponents directly. Let's rewrite both terms using base 3. Since $9 = 3^2$ and $27 = 3^3$ , we can substitute: $\left(9^{x}\right)^{1/2} = \left((3^2)^{x}\right)^{1/2} = (3^{2x})^{1/2} = 3^x$ $\left(27^{x}\right)^{1/3} = \left((3^3)^{x}\right)^{1/3} = (3^{3x})^{1/3} = 3^x$ Now our equation becomes: $3^x \times 3^x = 3^{10}$ Using the rule that $a^m \times a^n = a^{m+n}$ : $3^{x+x} = 3^{2x} = 3^{10}$ Since the bases are equal, the exponents must be equal: $2x = 10$ , so $x = 5$ . Let's examine why the other answers are incorrect: Choice A (4): If $x = 4$ , then $2x = 8$ , giving us $3^8$ , not $3^{10}$ . Choice C (6): If $x = 6$ , then $2x = 12$ , giving us $3^{12}$ , which is far too large. Choice D (10): If $x = 10$ , then $2x = 20$ , giving us $3^{20}$ , which is enormously larger than $3^{10}$ . Strategy tip: When dealing with exponential equations, always convert to a common base first. This transforms a complex-looking equation into a simple comparison of exponents, making the algebra much more manageable.
Find the integer $n$ such that $3^{n}<500<3^{n+1}$ .
Answer: 5 This question tests your ability to work with exponential inequalities and find bounds using powers of a base number. When you see an inequality like $3^{n}<500<3^{n+1}$ , you need to find which consecutive powers of 3 bracket the given value. To solve this, calculate powers of 3 systematically until you find where 500 falls. Start with smaller exponents and work up: $3^4 = 81$ $3^5 = 243$ $3^6 = 729$ Now compare these to 500. Since $3^5 = 243 < 500$ and $3^6 = 729 > 500$ , we have $243 < 500 < 729$ , which means $3^5 < 500 < 3^6$ . This matches the form $3^n < 500 < 3^{n+1}$ when $n = 5$ . Looking at the wrong answers: Choice (A) gives $n = 4$ , which would mean $3^4 < 500 < 3^5$ , or $81 < 500 < 243$ . This is impossible since 243 < 500. Choice (C) gives $n = 6$ , meaning $3^6 < 500 < 3^7$ , or $729 < 500 < 2187$ . This fails because 500 < 729. Choice (D) gives $n = 7$ , meaning $3^7 < 500 < 3^8$ , which also fails since $3^7 = 2187 > 500$ . Strategy tip: For exponential inequality problems, systematically calculate consecutive powers until you bracket the target value. Always verify both inequalities in your final answer—many students forget to check that both sides work.
What is the value of $\dfrac{16^{3/4}}{2^{1/2}}$ ?
Answer: $4\sqrt{2}$ When you encounter expressions with fractional exponents, the key is to convert everything to the same base and apply exponent rules systematically. Start by expressing both terms using base 2. Since $16 = 2^4$ , we can rewrite $16^{3/4}$ as $(2^4)^{3/4} = 2^{4 \cdot 3/4} = 2^3 = 8$ . The denominator $2^{1/2}$ stays as is. So our expression becomes: $\dfrac{8}{2^{1/2}} = \dfrac{8}{\sqrt{2}}$ To rationalize this, multiply both numerator and denominator by $\sqrt{2}$ : $\dfrac{8}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{8\sqrt{2}}{2} = 4\sqrt{2}$ This confirms answer choice A is correct. Let's examine why the other options miss the mark. Choice B ( $8\sqrt{2}$ ) represents what you'd get if you forgot to divide by the $\sqrt{2}$ factor entirely—essentially just calculating $16^{3/4}$ without the denominator. Choice C ( $\dfrac{4}{\sqrt{2}}$ ) appears if you correctly find the numerator as 8 but then incorrectly think the denominator contributes an extra factor of 2. Choice D ( $16\sqrt{2}$ ) likely results from miscalculating $16^{3/4}$ as 16 instead of 8. Strategy tip: When working with fractional exponents, always convert to a common base first, then apply the power rule $(a^m)^n = a^{mn}$ . This systematic approach prevents calculation errors and makes complex expressions much more manageable.
If $x^{2}-\dfrac{1}{x^{2}}=15$ and $x>1$ , what is the value of $x-\dfrac{1}{x}?$
Answer: $\sqrt{13}$ When you encounter algebraic expressions involving $x - \frac{1}{x}$ and $x^2 - \frac{1}{x^2}$ , look for the fundamental relationship between them. These expressions are connected through the algebraic identity: $(x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}$ . To find $x - \frac{1}{x}$ , rearrange this identity: $x^2 - \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2$ . Since we know $x^2 - \frac{1}{x^2} = 15$ , we can substitute: $15 = (x - \frac{1}{x})^2 + 2$ . Solving for $(x - \frac{1}{x})^2$ : $(x - \frac{1}{x})^2 = 13$ . Taking the square root: $x - \frac{1}{x} = \pm\sqrt{13}$ . Since $x > 1$ , we know that $x$ is positive and greater than 1, which means $\frac{1}{x} < 1$ . Therefore, $x - \frac{1}{x} > 0$ , so we take the positive square root: $x - \frac{1}{x} = \sqrt{13}$ . Answer choice A) $\sqrt{13}$ is correct. Choice B) 13 represents the value before taking the square root—a common error when students forget this final step. Choice C) 15 is the given value of $x^2 - \frac{1}{x^2}$ , which students might mistakenly think equals $x - \frac{1}{x}$ . Choice D) $\sqrt{15}$ results from incorrectly assuming $x - \frac{1}{x} = \sqrt{x^2 - \frac{1}{x^2}}$ without recognizing the algebraic relationship. Strategy tip: Memorize the identity $(a - b)^2 = a^2 - 2ab + b^2$ and its variations. These appear frequently on the GMAT when expressions involve reciprocals.
If $3^{2x} \cdot 9^{x-1} = 27^{x+1}$ , what is the value of $x$ ?
Answer: $x = 5$ First, express everything in terms of base 3: $3^{2x} \cdot (3^2)^{x-1} = (3^3)^{x+1}$ . This becomes $3^{2x} \cdot 3^{2(x-1)} = 3^{3(x+1)}$ , which simplifies to $3^{2x} \cdot 3^{2x-2} = 3^{3x+3}$ . Using the product rule: $3^{2x+2x-2} = 3^{3x+3}$ , so $3^{4x-2} = 3^{3x+3}$ . Since the bases are equal, the exponents must be equal: $4x-2 = 3x+3$ . Solving: $x = 5$ . Choice A results from solving $2x = x+3$ (forgetting the exponent rules). Choice C comes from $4x-2 = 2x+3$ (misreading $27$ as $3^2$ ). Choice D results from $4x-2 = x+11$ (computational error).
If $\sqrt[4]{16x^8} = 2x^n$ for all positive values of $x$ , what is the value of $n$ ?
Answer: $n = 2$ $\sqrt[4]{16x^8} = (16x^8)^{\frac{1}{4}} = 16^{\frac{1}{4}} \cdot (x^8)^{\frac{1}{4}} = (2^4)^{\frac{1}{4}} \cdot x^{8 \cdot \frac{1}{4}} = 2^{4 \cdot \frac{1}{4}} \cdot x^2 = 2^1 \cdot x^2 = 2x^2$ . Therefore $n = 2$ . Choice A results from incorrectly computing $x^{8 \cdot \frac{1}{4}} = x^1$ . Choice C comes from confusing the fourth root with the fourth power. Choice D results from not applying the fractional exponent to the variable's exponent.
If $0<p<1$ , which of the following expressions is always greater than 1?
Answer: $p^{-1}$ This question tests your understanding of how exponents behave when the base is a fraction between 0 and 1. When you see constraints like $0<p<1$ , immediately consider how different operations will transform values in this range. Let's examine what happens to each expression when $p$ is between 0 and 1. For option A, $p^{-1} = \frac{1}{p}$ . Since $p$ is a positive fraction less than 1, its reciprocal must be greater than 1. For example, if $p = 0.5$ , then $p^{-1} = \frac{1}{0.5} = 2$ . This relationship holds for any value between 0 and 1. Now let's see why the other options fail. Option B, $p^{2}$ , represents squaring a fraction. When you square any number between 0 and 1, the result becomes even smaller. For instance, $(0.5)^{2} = 0.25$ . Option C, $\sqrt{p}$ , takes the square root of a fraction between 0 and 1. While this makes the number larger than $p^{2}$ , it's still less than the original $p$ and therefore less than 1. For example, $\sqrt{0.25} = 0.5$ . Option D, $p^{1/3}$ , follows the same pattern as the square root—it increases the value compared to higher powers but keeps it below 1. Study tip: Remember that negative exponents create reciprocals, and the reciprocal of any proper fraction (between 0 and 1) is always greater than 1. This is a fundamental relationship that appears frequently on the GMAT.
Suppose $\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers with $a>b$ . What is the value of $a-b$ ?
Answer: 1 When you encounter nested radicals like $\sqrt{5+2\sqrt{6}}$ , the key insight is to express them in the form $\sqrt{a}+\sqrt{b}$ by working backwards. This technique involves squaring both sides and using algebraic manipulation. Let's assume $\sqrt{5+2\sqrt{6}} = \sqrt{a}+\sqrt{b}$ and square both sides: $5+2\sqrt{6} = (\sqrt{a}+\sqrt{b})^2 = a + b + 2\sqrt{ab}$ For this equation to hold, the rational and irrational parts must match separately: - Rational parts: $5 = a + b$ - Irrational parts: $2\sqrt{6} = 2\sqrt{ab}$ From the irrational parts, we get $\sqrt{6} = \sqrt{ab}$ , so $ab = 6$ . Now you have a system: $a + b = 5$ and $ab = 6$ . These are the sum and product of two numbers, which means $a$ and $b$ are roots of the quadratic $x^2 - 5x + 6 = 0$ . Factoring gives $(x-2)(x-3) = 0$ , so the solutions are 2 and 3. Since $a > b$ , we have $a = 3$ and $b = 2$ . Therefore, $a - b = 3 - 2 = 1$ . Choice A is correct. Choice B (2) would result if you mistakenly used $a = 4, b = 2$ or similar incorrect pairs. Choice C (3) occurs if you confused which value represents $a-b$ and selected just $a$ . Choice D (4) might arise from calculation errors in the quadratic factorization. Strategy tip: When denesting radicals, always square both sides and separate rational from irrational components. The resulting system often leads to a simple quadratic equation.

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