GMAT Quantitative Quiz: Exponents And Roots
17 questions · exam conditions
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Exponents And RootsQuestion 1 of 17

If 32x9x1=27x+13^{2x} \cdot 9^{x-1} = 27^{x+1}, what is the value of xx?

x=3x = 3
x=5x = 5
x=7x = 7
x=9x = 9
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GMAT Quantitative Quiz

GMAT Quantitative Quiz: Exponents And Roots

Practice Exponents And Roots in GMAT 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 And Roots, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT 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 32x9x1=27x+13^{2x} \cdot 9^{x-1} = 27^{x+1}, what is the value of xx?

  1. x=3x = 3
  2. x=5x = 5 (correct answer)
  3. x=7x = 7
  4. x=9x = 9

Explanation: First, express everything in terms of base 3: 32x(32)x1=(33)x+13^{2x} \cdot (3^2)^{x-1} = (3^3)^{x+1}. This becomes 32x32(x1)=33(x+1)3^{2x} \cdot 3^{2(x-1)} = 3^{3(x+1)}, which simplifies to 32x32x2=33x+33^{2x} \cdot 3^{2x-2} = 3^{3x+3}. Using the product rule: 32x+2x2=33x+33^{2x+2x-2} = 3^{3x+3}, so 34x2=33x+33^{4x-2} = 3^{3x+3}. Since the bases are equal, the exponents must be equal: 4x2=3x+34x-2 = 3x+3. Solving: x=5x = 5. Choice A results from solving 2x=x+32x = x+3 (forgetting the exponent rules). Choice C comes from 4x2=2x+34x-2 = 2x+3 (misreading 2727 as 323^2). Choice D results from 4x2=x+114x-2 = x+11 (computational error).

Question 2

If 16x84=2xn\sqrt[4]{16x^8} = 2x^n for all positive values of xx, what is the value of nn?

  1. n=1n = 1
  2. n=2n = 2 (correct answer)
  3. n=4n = 4
  4. n=8n = 8

Explanation: 16x84=(16x8)14=1614(x8)14=(24)14x814=2414x2=21x2=2x2\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=2n = 2. Choice A results from incorrectly computing x814=x1x^{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.

Question 3

If x21x2=15x^{2}-\dfrac{1}{x^{2}}=15 and x>1x>1, what is the value of x1x?x-\dfrac{1}{x}?

  1. 13\sqrt{13} (correct answer)
  2. 13
  3. 15
  4. 15\sqrt{15}

Explanation: When you encounter algebraic expressions involving x1xx - \frac{1}{x} and x21x2x^2 - \frac{1}{x^2}, look for the fundamental relationship between them. These expressions are connected through the algebraic identity: (x1x)2=x22+1x2(x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}. To find x1xx - \frac{1}{x}, rearrange this identity: x21x2=(x1x)2+2x^2 - \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2. Since we know x21x2=15x^2 - \frac{1}{x^2} = 15, we can substitute: 15=(x1x)2+215 = (x - \frac{1}{x})^2 + 2. Solving for (x1x)2(x - \frac{1}{x})^2: (x1x)2=13(x - \frac{1}{x})^2 = 13. Taking the square root: x1x=±13x - \frac{1}{x} = \pm\sqrt{13}. Since x>1x > 1, we know that xx is positive and greater than 1, which means 1x<1\frac{1}{x} < 1. Therefore, x1x>0x - \frac{1}{x} > 0, so we take the positive square root: x1x=13x - \frac{1}{x} = \sqrt{13}. Answer choice A) 13\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 x21x2x^2 - \frac{1}{x^2}, which students might mistakenly think equals x1xx - \frac{1}{x}. Choice D) 15\sqrt{15} results from incorrectly assuming x1x=x21x2x - \frac{1}{x} = \sqrt{x^2 - \frac{1}{x^2}} without recognizing the algebraic relationship. Strategy tip: Memorize the identity (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2 and its variations. These appear frequently on the GMAT when expressions involve reciprocals.

Question 4

Which of the following equals 324\sqrt[4]{32} ?

  1. 21.22^{1.2}
  2. 2242\sqrt[4]{2} (correct answer)
  3. 4244\sqrt[4]{2}
  4. 8\sqrt{8}

Explanation: 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 324\sqrt[4]{32}, start by expressing 32 as a power of 2: 32=2532 = 2^5. So 324=254=(25)1/4=25/4\sqrt[4]{32} = \sqrt[4]{2^5} = (2^5)^{1/4} = 2^{5/4}. Now rewrite this exponent: 25/4=21+1/4=2121/4=221/42^{5/4} = 2^{1 + 1/4} = 2^1 \cdot 2^{1/4} = 2 \cdot 2^{1/4}. Since 21/4=242^{1/4} = \sqrt[4]{2}, we get 2242\sqrt[4]{2}, which is choice B. Choice A gives 21.2=26/52^{1.2} = 2^{6/5}, which doesn't equal 25/42^{5/4} since 65=2420\frac{6}{5} = \frac{24}{20} while 54=2520\frac{5}{4} = \frac{25}{20}. Choice C gives 424=2221/4=22+1/4=29/44\sqrt[4]{2} = 2^2 \cdot 2^{1/4} = 2^{2 + 1/4} = 2^{9/4}, which is too large. Choice D gives 8=23=23/2\sqrt{8} = \sqrt{2^3} = 2^{3/2}. Converting to compare: 32=64\frac{3}{2} = \frac{6}{4} while our answer is 25/42^{5/4}, so this is also incorrect. Strategy tip: When simplifying radicals, always convert to exponential form first (an=a1/n\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.

Question 5

For non-zero aa, the expression a3a5a4\dfrac{a^{3}\,a^{-5}}{a^{-4}} can be written in the form aka^{k}. What is the value of kk?

  1. -6
  2. -2
  3. 2 (correct answer)
  4. 12

Explanation: 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 a3a5a4\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: a3a5=a3+(5)=a2a^{3} \cdot a^{-5} = a^{3+(-5)} = a^{-2}. Now you have a2a4\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: a2÷a4=a2(4)=a2+4=a2a^{-2} \div a^{-4} = a^{-2-(-4)} = a^{-2+4} = a^{2}. Therefore, k=2k = 2, making C correct. Looking at the wrong answers: A) -6 likely comes from incorrectly adding all the exponents: 3+(5)+(4)=63 + (-5) + (-4) = -6, but this ignores the division operation. B) -2 represents stopping after combining the numerator (a3a5=a2a^{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=2a = 2.

Question 6

If 8x63=2xm\sqrt[3]{8x^{6}}=2x^{m} for all positive real xx, what is the value of mm?

  1. 1/3
  2. 3
  3. 2 (correct answer)
  4. 6

Explanation: 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 8x63=2xm\sqrt[3]{8x^{6}}=2x^{m}. The cube root can be written as a fractional exponent: 8x63=(8x6)1/3\sqrt[3]{8x^{6}} = (8x^{6})^{1/3}. Breaking this down further: (8x6)1/3=81/3(x6)1/3(8x^{6})^{1/3} = 8^{1/3} \cdot (x^{6})^{1/3}. Since 8=238 = 2^3, we have 81/3=(23)1/3=231/3=21=28^{1/3} = (2^3)^{1/3} = 2^{3 \cdot 1/3} = 2^1 = 2. For the variable part, (x6)1/3=x61/3=x2(x^{6})^{1/3} = x^{6 \cdot 1/3} = x^2. Therefore, 8x63=2x2\sqrt[3]{8x^{6}} = 2x^2, which means 2x2=2xm2x^2 = 2x^{m}. Since the coefficients are equal and this must hold for all positive real xx, we need x2=xmx^2 = x^{m}, so m=2m = 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 mm. Choice (D) 6 might appeal if you forgot to apply the fractional exponent rule and just used the original exponent from x6x^6. When working with radicals and exponents, always convert radicals to fractional exponents first, then apply the power rule (am)n=amn(a^m)^n = a^{mn} systematically to avoid calculation errors.

Question 7

For positive xx, (1x3)1/2=xk\left(\dfrac{1}{x^{3}}\right)^{\,1/2}=x^{k}. What is kk?

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

Explanation: 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 (1x3)1/2\left(\dfrac{1}{x^{3}}\right)^{\,1/2}, first rewrite the fraction using negative exponents: 1x3=x3\dfrac{1}{x^{3}} = x^{-3}. This gives us (x3)1/2(x^{-3})^{1/2}. Now apply the power rule: when raising a power to another power, multiply the exponents. So (x3)1/2=x(3)(1/2)=x3/2(x^{-3})^{1/2} = x^{(-3) \cdot (1/2)} = x^{-3/2}. Since this equals xkx^k, we have k=32k = -\dfrac{3}{2}. Looking at the wrong answers: Choice (A) gives k=3k = -3, which would result from forgetting to apply the outer exponent of 12\frac{1}{2}. Choice (C) gives k=32k = \frac{3}{2}, which reflects correctly applying the exponent rules but forgetting that 1x3=x3\frac{1}{x^3} = x^{-3}, not x3x^3. Choice (D) gives k=23k = \frac{2}{3}, which comes from incorrectly multiplying the exponents as 123=32\frac{1}{2} \cdot 3 = \frac{3}{2} but then inverting it, while also missing the negative sign. The correct answer is (B) k=32k = -\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.

Question 8

What is the value of 163/421/2\dfrac{16^{3/4}}{2^{1/2}}?

  1. 424\sqrt{2} (correct answer)
  2. 828\sqrt{2}
  3. 42\dfrac{4}{\sqrt{2}}
  4. 16216\sqrt{2}

Explanation: 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=2416 = 2^4, we can rewrite 163/416^{3/4} as (24)3/4=243/4=23=8(2^4)^{3/4} = 2^{4 \cdot 3/4} = 2^3 = 8. The denominator 21/22^{1/2} stays as is. So our expression becomes: 821/2=82\dfrac{8}{2^{1/2}} = \dfrac{8}{\sqrt{2}} To rationalize this, multiply both numerator and denominator by 2\sqrt{2}: 8222=822=42\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 (828\sqrt{2}) represents what you'd get if you forgot to divide by the 2\sqrt{2} factor entirely—essentially just calculating 163/416^{3/4} without the denominator. Choice C (42\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 (16216\sqrt{2}) likely results from miscalculating 163/416^{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 (am)n=amn(a^m)^n = a^{mn}. This systematic approach prevents calculation errors and makes complex expressions much more manageable.

Question 9

Positive numbers r,s,tr, s, t satisfy r2=s3=t5=kr^{2}=s^{3}=t^{5}=k. Which of the following is equal to k1/30k^{\,1/30}?

  1. r1/15r^{1/15} (correct answer)
  2. s1/12s^{1/12}
  3. t1/3t^{1/3}
  4. r1/10r^{1/10}

Explanation: 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 r2=s3=t5=kr^{2}=s^{3}=t^{5}=k, you can express each variable in terms of kk. From r2=kr^{2}=k, you get r=k1/2r=k^{1/2}. From s3=ks^{3}=k, you get s=k1/3s=k^{1/3}. From t5=kt^{5}=k, you get t=k1/5t=k^{1/5}. To find k1/30k^{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: r1/15=(k1/2)1/15=k1/30r^{1/15} = (k^{1/2})^{1/15} = k^{1/30}. This matches exactly what we're looking for. Choice B: s1/12=(k1/3)1/12=k1/36s^{1/12} = (k^{1/3})^{1/12} = k^{1/36}. This gives the wrong exponent. Choice C: t1/3=(k1/5)1/3=k1/15t^{1/3} = (k^{1/5})^{1/3} = k^{1/15}. This exponent is too large. Choice D: r1/10=(k1/2)1/10=k1/20r^{1/10} = (k^{1/2})^{1/10} = k^{1/20}. This exponent is also too large. Only choice A produces the required exponent of 1/301/30. 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.

Question 10

If (9x)1/2×(27x)1/3=310\left(9^{x}\right)^{1/2}\,\times\,\left(27^{x}\right)^{1/3}=3^{10}, what is the value of xx?

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 10

Explanation: 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=329 = 3^2 and 27=3327 = 3^3, we can substitute: (9x)1/2=((32)x)1/2=(32x)1/2=3x\left(9^{x}\right)^{1/2} = \left((3^2)^{x}\right)^{1/2} = (3^{2x})^{1/2} = 3^x (27x)1/3=((33)x)1/3=(33x)1/3=3x\left(27^{x}\right)^{1/3} = \left((3^3)^{x}\right)^{1/3} = (3^{3x})^{1/3} = 3^x Now our equation becomes: 3x×3x=3103^x \times 3^x = 3^{10} Using the rule that am×an=am+na^m \times a^n = a^{m+n}: 3x+x=32x=3103^{x+x} = 3^{2x} = 3^{10} Since the bases are equal, the exponents must be equal: 2x=102x = 10, so x=5x = 5. Let's examine why the other answers are incorrect: Choice A (4): If x=4x = 4, then 2x=82x = 8, giving us 383^8, not 3103^{10}. Choice C (6): If x=6x = 6, then 2x=122x = 12, giving us 3123^{12}, which is far too large. Choice D (10): If x=10x = 10, then 2x=202x = 20, giving us 3203^{20}, which is enormously larger than 3103^{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.

Question 11

Suppose 5+26=a+b\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b} where aa and bb are positive integers with a>ba>b. What is the value of aba-b?

  1. 1 (correct answer)
  2. 2
  3. 3
  4. 4

Explanation: When you encounter nested radicals like 5+26\sqrt{5+2\sqrt{6}}, the key insight is to express them in the form a+b\sqrt{a}+\sqrt{b} by working backwards. This technique involves squaring both sides and using algebraic manipulation. Let's assume 5+26=a+b\sqrt{5+2\sqrt{6}} = \sqrt{a}+\sqrt{b} and square both sides: 5+26=(a+b)2=a+b+2ab5+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+b5 = a + b
  • Irrational parts: 26=2ab2\sqrt{6} = 2\sqrt{ab}
From the irrational parts, we get 6=ab\sqrt{6} = \sqrt{ab}, so ab=6ab = 6. Now you have a system: a+b=5a + b = 5 and ab=6ab = 6. These are the sum and product of two numbers, which means aa and bb are roots of the quadratic x25x+6=0x^2 - 5x + 6 = 0. Factoring gives (x2)(x3)=0(x-2)(x-3) = 0, so the solutions are 2 and 3. Since a>ba > b, we have a=3a = 3 and b=2b = 2. Therefore, ab=32=1a - b = 3 - 2 = 1. Choice A is correct. Choice B (2) would result if you mistakenly used a=4,b=2a = 4, b = 2 or similar incorrect pairs. Choice C (3) occurs if you confused which value represents aba-b and selected just aa. 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.

Question 12

Find the integer nn such that 3n<500<3n+13^{n}<500<3^{n+1}.

  1. 4
  2. 5 (correct answer)
  3. 6
  4. 7

Explanation: 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 3n<500<3n+13^{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: 34=813^4 = 81 35=2433^5 = 243
36=7293^6 = 729
Now compare these to 500. Since 35=243<5003^5 = 243 < 500 and 36=729>5003^6 = 729 > 500, we have 243<500<729243 < 500 < 729, which means 35<500<363^5 < 500 < 3^6. This matches the form 3n<500<3n+13^n < 500 < 3^{n+1} when n=5n = 5. Looking at the wrong answers: Choice (A) gives n=4n = 4, which would mean 34<500<353^4 < 500 < 3^5, or 81<500<24381 < 500 < 243. This is impossible since 243 < 500. Choice (C) gives n=6n = 6, meaning 36<500<373^6 < 500 < 3^7, or 729<500<2187729 < 500 < 2187. This fails because 500 < 729. Choice (D) gives n=7n = 7, meaning 37<500<383^7 < 500 < 3^8, which also fails since 37=2187>5003^7 = 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.

Question 13

If 0<p<10<p<1, which of the following expressions is always greater than 1?

  1. p1p^{-1} (correct answer)
  2. p2p^{2}
  3. p\sqrt{p}
  4. p1/3p^{1/3}

Explanation: 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<10<p<1, immediately consider how different operations will transform values in this range. Let's examine what happens to each expression when pp is between 0 and 1. For option A, p1=1pp^{-1} = \frac{1}{p}. Since pp is a positive fraction less than 1, its reciprocal must be greater than 1. For example, if p=0.5p = 0.5, then p1=10.5=2p^{-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, p2p^{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(0.5)^{2} = 0.25. Option C, p\sqrt{p}, takes the square root of a fraction between 0 and 1. While this makes the number larger than p2p^{2}, it's still less than the original pp and therefore less than 1. For example, 0.25=0.5\sqrt{0.25} = 0.5. Option D, p1/3p^{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.

Question 14

For positive real numbers kk and nn satisfying 2k=10n2^{k}=10^{n}, which of the following expresses nn in terms of kk?

  1. klog102k\log_{10}2 (correct answer)
  2. klog210k\log_{2}10
  3. klog210\dfrac{k}{\log_{2}10}
  4. klog102\dfrac{k}{\log_{10}2}

Explanation: When you encounter equations with different bases like 2k=10n2^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 2k=10n2^k = 10^n, take the logarithm base 10 of both sides: log10(2k)=log10(10n)\log_{10}(2^k) = \log_{10}(10^n) Using the logarithm power rule, logb(xa)=alogb(x)\log_b(x^a) = a\log_b(x): klog10(2)=nlog10(10)k \cdot \log_{10}(2) = n \cdot \log_{10}(10) Since log10(10)=1\log_{10}(10) = 1: klog10(2)=nk \cdot \log_{10}(2) = n Therefore, n=klog102n = k\log_{10}2, which is choice (A). Let's examine why the other options are incorrect: Choice (B) gives klog210k\log_2 10. This would be correct if we had solved for kk in terms of nn by taking log2\log_2 of both sides, but we need nn in terms of kk. Choice (C) gives klog210\frac{k}{\log_2 10}. This represents a common algebraic error—incorrectly manipulating the relationship between klog210k\log_2 10 and the desired form. Choice (D) gives klog102\frac{k}{\log_{10} 2}. This might seem appealing because it contains log102\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 nn (related to base 10), using log10\log_{10} makes the algebra cleanest.

Question 15

If x4+8x2+16=x2+4\sqrt{x^4 + 8x^2 + 16} = x^2 + 4 for all real values of xx, which of the following must be true?

  1. x0x \geq 0 for all solutions
  2. x2+40x^2 + 4 \geq 0 for all solutions
  3. The equation has no real solutions
  4. The equation is satisfied by all real numbers (correct answer)

Explanation: Notice that x4+8x2+16=(x2+4)2x^4 + 8x^2 + 16 = (x^2 + 4)^2. Therefore, x4+8x2+16=(x2+4)2=x2+4\sqrt{x^4 + 8x^2 + 16} = \sqrt{(x^2 + 4)^2} = |x^2 + 4|. Since x20x^2 \geq 0 for all real xx, we have x2+44>0x^2 + 4 \geq 4 > 0, so x2+4=x2+4|x^2 + 4| = x^2 + 4. Thus the equation becomes x2+4=x2+4x^2 + 4 = x^2 + 4, which is true for all real xx. Choice A is incorrect because xx 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.

Question 16

If x+x+x+=5\sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} = 5 for x>0x > 0, what is the value of xx?

  1. x=20x = 20 (correct answer)
  2. x=25x = 25
  3. x=30x = 30
  4. x=35x = 35

Explanation: Let y=x+x+x+y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}. Since this is an infinite nested radical that converges, we have y=x+yy = \sqrt{x + y}. Given that y=5y = 5, we get 5=x+55 = \sqrt{x + 5}. Squaring both sides: 25=x+525 = x + 5, so x=20x = 20. We can verify: if x=20x = 20, then y=20+yy = \sqrt{20 + y}, and with y=5y = 5, we get 5=20+5=25=55 = \sqrt{20 + 5} = \sqrt{25} = 5 ✓. Choice B comes from incorrectly setting up y2=xy^2 = x instead of y2=x+yy^2 = x + y. Choice C results from 52+5=305^2 + 5 = 30 (adding instead of subtracting). Choice D comes from 5×7=355 \times 7 = 35 (using an incorrect relationship).

Question 17

If 2x4y8z3=16\sqrt[3]{2^x \cdot 4^y \cdot 8^z} = 16, what is the value of x+2y+3zx + 2y + 3z?

  1. 2424
  2. 1616
  3. 1212 (correct answer)
  4. 3636

Explanation: 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=224 = 2^2 and 8=238 = 2^3. So the equation becomes: 2x(22)y(23)z3=16\sqrt[3]{2^x \cdot (2^2)^y \cdot (2^3)^z} = 16 Using the power rule (am)n=amn(a^m)^n = a^{mn}, this simplifies to: 2x22y23z3=16\sqrt[3]{2^x \cdot 2^{2y} \cdot 2^{3z}} = 16 When multiplying powers with the same base, add the exponents: 2x+2y+3z3=16\sqrt[3]{2^{x+2y+3z}} = 16 A cube root can be written as a fractional exponent: 2(x+2y+3z)/3=162^{(x+2y+3z)/3} = 16 Since 16=2416 = 2^4, you have: 2(x+2y+3z)/3=242^{(x+2y+3z)/3} = 2^4 When the bases are equal, the exponents must be equal: x+2y+3z3=4\frac{x+2y+3z}{3} = 4 Therefore: x+2y+3z=12x+2y+3z = 12 The answer is C) 1212. Let's examine why the other choices are wrong. Choice A) 2424 would result if you incorrectly multiplied by 3 instead of dividing when solving the final equation. Choice B) 1616 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) 3636 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.