ACCUPLACER Advanced Algebra & Functions Quiz: Solving Exponential Equations
20 questions · exam conditions
0:00
Solving Exponential EquationsQuestion 1 of 20

If 5x⋅25x=1255^{x} \cdot 25^{x} = 125, what is the value of xx?

13\frac{1}{3}
23\frac{2}{3}
1
3
← Back to quizzes

ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Exponential Equations

Practice Solving Exponential Equations in ACCUPLACER Advanced Algebra & Functions 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 Solving Exponential Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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 5x⋅25x=1255^{x} \cdot 25^{x} = 125, what is the value of xx?

  1. 13\frac{1}{3}
  2. 23\frac{2}{3}
  3. 1 (correct answer)
  4. 3
Explanation: First, express all terms with a base of 5. 25=5225 = 5^2 and 125=53125 = 5^3. The equation becomes 5x⋅(52)x=535^x \cdot (5^2)^x = 5^3. This simplifies to 5x⋅52x=535^x \cdot 5^{2x} = 5^3. Using the product rule for exponents, add the exponents on the left side: 5x+2x=535^{x+2x} = 5^3, or 53x=535^{3x} = 5^3. Equating the exponents gives 3x=33x = 3, so x=1x=1.

Question 2

The equation 27x=9x+227^{x} = 9^{x+2} has a solution of the form x=kx=k. What is the value of 2k−12k-1?

  1. 3
  2. 4
  3. 6
  4. 7 (correct answer)
Explanation: First, solve for xx by finding a common base for 27 and 9, which is 3. 27=3327 = 3^3 and 9=329 = 3^2. The equation becomes (33)x=(32)x+2(3^3)^x = (3^2)^{x+2}. This simplifies to 33x=32(x+2)3^{3x} = 3^{2(x+2)}, or 33x=32x+43^{3x} = 3^{2x+4}. Equating the exponents gives 3x=2x+43x = 2x+4. Solving for xx gives x=4x=4. So, k=4k=4. The question asks for the value of 2k−12k-1. Substitute k=4k=4: 2(4)−1=8−1=72(4)-1 = 8-1 = 7.

Question 3

If 12⋅8x=16\frac{1}{2} \cdot 8^{x} = 16, what is the value of xx?

  1. 43\frac{4}{3}
  2. 53\frac{5}{3} (correct answer)
  3. 2
  4. 3
Explanation: First, isolate the exponential term 8x8^x by multiplying both sides of the equation by 2: 8x=16×2=328^x = 16 \times 2 = 32. Now, express both 8 and 32 as powers of a common base, which is 2. 8=238 = 2^3 and 32=2532 = 2^5. The equation becomes (23)x=25(2^3)^x = 2^5, which simplifies to 23x=252^{3x} = 2^5. Equating the exponents gives 3x=53x = 5, so x=53x = \frac{5}{3}.

Question 4

What is the solution to the equation 92x=279^{2x} = 27?

  1. 13\frac{1}{3}
  2. 34\frac{3}{4} (correct answer)
  3. 32\frac{3}{2}
  4. 3
Explanation: Express both sides of the equation with a common base of 3. 9=329 = 3^2 and 27=3327 = 3^3. The equation becomes (32)2x=33(3^2)^{2x} = 3^3. Simplify the left side to get 34x=333^{4x} = 3^3. Now, set the exponents equal to each other: 4x=34x = 3. Solving for xx gives x=34x = \frac{3}{4}.

Question 5

The equation 23x−1=4x+22^{3x-1} = 4^{x+2} has solution x=kx = k. What is the value of 3k3k?

  1. 3k=153k = 15 (correct answer)
  2. 3k=123k = 12
  3. 3k=183k = 18
  4. 3k=213k = 21
Explanation: Rewrite with base 2: 4x+2=(22)x+2=22(x+2)=22x+44^{x+2} = (2^2)^{x+2} = 2^{2(x+2)} = 2^{2x+4}. The equation becomes 23x−1=22x+42^{3x-1} = 2^{2x+4}. Setting exponents equal: 3x−1=2x+43x-1 = 2x+4, so x=5x = 5. Therefore 3k=3(5)=153k = 3(5) = 15. Choice B results from solving 3x−1=2x+33x-1 = 2x+3 (arithmetic error). Choice C comes from incorrectly getting x=6x = 6. Choice D results from solving 3x−1=2x+53x-1 = 2x+5 (error in expanding the right side).

Question 6

If 7x−3=497^{x-3} = 49, what is the value of x2−xx^2 - x?

  1. 2
  2. 5
  3. 20 (correct answer)
  4. 90
Explanation: First, solve for xx. Rewrite 49 as 727^2. The equation is 7x−3=727^{x-3} = 7^2. Equating the exponents gives x−3=2x-3 = 2, so x=5x=5. The question asks for the value of x2−xx^2 - x. Substitute x=5x=5 into the expression: 52−5=25−5=205^2 - 5 = 25 - 5 = 20.

Question 7

What is the solution to the equation 16x−1=32x16^{x-1} = 32^x?

  1. -4 (correct answer)
  2. -1
  3. 1
  4. 4
Explanation: To solve, find a common base for 16 and 32, which is 2. 16=2416 = 2^4 and 32=2532 = 2^5. The equation becomes (24)x−1=(25)x(2^4)^{x-1} = (2^5)^x. Apply the power rule for exponents: 24(x−1)=25x2^{4(x-1)} = 2^{5x}, which simplifies to 24x−4=25x2^{4x-4} = 2^{5x}. Now, equate the exponents: 4x−4=5x4x-4 = 5x. Subtracting 4x4x from both sides gives −4=x-4 = x.

Question 8

How many distinct real solutions exist for the equation 3x2+2x=1273^{x^2+2x} = \frac{1}{27}?

  1. 0 (correct answer)
  2. 1
  3. 2
  4. 3
Explanation: First, write both sides with a base of 3. 127=3−3\frac{1}{27} = 3^{-3}. The equation becomes 3x2+2x=3−33^{x^2+2x} = 3^{-3}. Equating the exponents gives the quadratic equation x2+2x=−3x^2+2x = -3, which can be rewritten as x2+2x+3=0x^2+2x+3=0. To find the number of real solutions, we check the discriminant, b2−4acb^2-4ac. Here, a=1,b=2,c=3a=1, b=2, c=3. The discriminant is 22−4(1)(3)=4−12=−82^2 - 4(1)(3) = 4 - 12 = -8. Since the discriminant is negative, there are no real solutions to the quadratic equation.

Question 9

For what value of kk is x=3x=3 a solution to the equation 9x−1=3k9^{x-1} = 3^{k}?

  1. 2
  2. 4 (correct answer)
  3. 5
  4. 6
Explanation: Substitute x=3x=3 into the equation: 93−1=3k9^{3-1} = 3^k, which simplifies to 92=3k9^2 = 3^k. Since 9=329 = 3^2, we can rewrite the left side as (32)2=3k(3^2)^2 = 3^k. This simplifies to 34=3k3^4 = 3^k. Therefore, k=4k=4.

Question 10

What is the product of the solutions to the equation 4x2=210x−124^{x^2} = 2^{10x-12}?

  1. -6
  2. -5
  3. 5
  4. 6 (correct answer)
Explanation: First, establish a common base of 2. Since 4=224 = 2^2, the equation can be rewritten as (22)x2=210x−12(2^2)^{x^2} = 2^{10x-12}. This simplifies to 22x2=210x−122^{2x^2} = 2^{10x-12}. Equating the exponents gives the quadratic equation 2x2=10x−122x^2 = 10x-12. Rearrange the equation to 2x2−10x+12=02x^2 - 10x + 12 = 0. Divide the entire equation by 2 to simplify it: x2−5x+6=0x^2 - 5x + 6 = 0. This factors to (x−2)(x−3)=0(x-2)(x-3) = 0. The solutions are x=2x=2 and x=3x=3. The product of the solutions is 2×3=62 \times 3 = 6.

Question 11

If 51−x=11255^{1-x} = \frac{1}{125}, what is the value of xx?

  1. -4
  2. -2
  3. 2
  4. 4 (correct answer)
Explanation: First, express 1125\frac{1}{125} as a power of 5. Since 125=53125 = 5^3, we have 1125=5−3\frac{1}{125} = 5^{-3}. The equation is now 51−x=5−35^{1-x} = 5^{-3}. Since the bases are equal, the exponents must be equal: 1−x=−31-x = -3. Subtracting 1 from both sides gives −x=−4-x = -4, so x=4x = 4.

Question 12

What is the solution to the equation 3⋅2x+1−5=433 \cdot 2^{x+1} - 5 = 43?

  1. 2
  2. 3 (correct answer)
  3. 4
  4. 5
Explanation: First, isolate the exponential term. Add 5 to both sides: 3⋅2x+1=483 \cdot 2^{x+1} = 48. Then, divide by 3: 2x+1=162^{x+1} = 16. Now, express 16 as a power of 2: 16=2416 = 2^4. The equation becomes 2x+1=242^{x+1} = 2^4. Set the exponents equal: x+1=4x+1 = 4. Solving for xx gives x=3x = 3.

Question 13

If 2x2−5=182^{x^2-5} = \frac{1}{8}, which of the following is a possible value of xx?

  1. 2\sqrt{2} (correct answer)
  2. 2
  3. 222\sqrt{2}
  4. 4
Explanation: Rewrite 18\frac{1}{8} as a power of 2, which is 2−32^{-3}. The equation becomes 2x2−5=2−32^{x^2-5} = 2^{-3}. Equating the exponents gives x2−5=−3x^2 - 5 = -3. Add 5 to both sides to get x2=2x^2 = 2. Taking the square root of both sides gives x=±2x = \pm\sqrt{2}. Therefore, 2\sqrt{2} is a possible value for xx.

Question 14

If 4x+1=8x−14^{x+1} = 8^{x-1}, what is the value of xx?

  1. 1
  2. 3
  3. 5 (correct answer)
  4. 7
Explanation: To solve the equation, rewrite both sides with a common base, which is 2. The equation becomes (22)x+1=(23)x−1(2^2)^{x+1} = (2^3)^{x-1}. Using the power of a power rule for exponents, this simplifies to 22(x+1)=23(x−1)2^{2(x+1)} = 2^{3(x-1)}, which is 22x+2=23x−32^{2x+2} = 2^{3x-3}. Since the bases are equal, the exponents must be equal: 2x+2=3x−32x+2 = 3x-3. Solving for xx gives 5=x5 = x.

Question 15

If 5x+1=25\sqrt{5}^{x+1} = 25, what is the value of xx?

  1. -3
  2. 1
  3. 3 (correct answer)
  4. 4
Explanation: Express both sides with a common base of 5. 5=51/2\sqrt{5} = 5^{1/2} and 25=5225 = 5^2. The equation becomes (51/2)x+1=52(5^{1/2})^{x+1} = 5^2. Simplify the left side to get 512(x+1)=525^{\frac{1}{2}(x+1)} = 5^2. Set the exponents equal: 12(x+1)=2\frac{1}{2}(x+1) = 2. Multiply both sides by 2 to get x+1=4x+1 = 4. Solving for xx gives x=3x = 3.

Question 16

If 82x−1=28^{2x-1} = \sqrt{2}, what is the value of xx?

  1. 14\frac{1}{4}
  2. 13\frac{1}{3}
  3. 712\frac{7}{12} (correct answer)
  4. 56\frac{5}{6}
Explanation: First, express both sides of the equation with a common base of 2. We have 8=238 = 2^3 and 2=21/2\sqrt{2} = 2^{1/2}. Substituting these into the equation gives (23)2x−1=21/2(2^3)^{2x-1} = 2^{1/2}. Using the power of a power rule, this becomes 23(2x−1)=21/22^{3(2x-1)} = 2^{1/2}, or 26x−3=21/22^{6x-3} = 2^{1/2}. Now, set the exponents equal: 6x−3=126x-3 = \frac{1}{2}. Add 3 to both sides: 6x=12+3=726x = \frac{1}{2} + 3 = \frac{7}{2}. Finally, divide by 6: x=7/26=712x = \frac{7/2}{6} = \frac{7}{12}.

Question 17

What is the sum of all possible solutions for xx in the equation 10x2−9=110^{x^2-9} = 1?

  1. -3
  2. 0 (correct answer)
  3. 3
  4. 6
Explanation: For any non-zero base bb, the equation by=1b^y = 1 implies that y=0y=0. In this problem, the exponent must be zero: x2−9=0x^2 - 9 = 0. This is a difference of squares, which factors to (x−3)(x+3)=0(x-3)(x+3) = 0. The solutions are x=3x=3 and x=−3x=-3. The sum of these solutions is 3+(−3)=03 + (-3) = 0.

Question 18

What is the solution to the equation 2x+4=4x+12^{x+4} = 4^{x+1}?

  1. 12\frac{1}{2}
  2. 1
  3. 2 (correct answer)
  4. 3
Explanation: To solve this equation, rewrite both sides with the same base, which is 2. Since 4=224 = 2^2, the equation becomes 2x+4=(22)x+12^{x+4} = (2^2)^{x+1}. Using the power of a power rule on the right side gives 2x+4=22(x+1)2^{x+4} = 2^{2(x+1)}, which simplifies to 2x+4=22x+22^{x+4} = 2^{2x+2}. Now that the bases are equal, the exponents must be equal: x+4=2x+2x+4 = 2x+2. Subtract xx from both sides: 4=x+24 = x+2. Subtract 2 from both sides: 2=x2 = x.

Question 19

If (49)x=278(\frac{4}{9})^{x} = \frac{27}{8}, what is the value of xx?

  1. −32-\frac{3}{2} (correct answer)
  2. −23-\frac{2}{3}
  3. 23\frac{2}{3}
  4. 32\frac{3}{2}
Explanation: Express both sides using common bases. 49=(23)2\frac{4}{9} = (\frac{2}{3})^2 and 278=(32)3\frac{27}{8} = (\frac{3}{2})^3. The equation becomes ((23)2)x=(32)3((\frac{2}{3})^2)^x = (\frac{3}{2})^3, which simplifies to (23)2x=(32)3(\frac{2}{3})^{2x} = (\frac{3}{2})^3. To make the bases the same, take the reciprocal of the right side and negate the exponent: (23)2x=(23)−3(\frac{2}{3})^{2x} = (\frac{2}{3})^{-3}. Now, equate the exponents: 2x=−32x = -3. Solving for xx gives x=−32x = -\frac{3}{2}.

Question 20

What is the solution to (13)x−2=81(\frac{1}{3})^{x-2} = 81?

  1. -6
  2. -2 (correct answer)
  3. 4
  4. 6
Explanation: Rewrite both sides with a common base of 3. 13=3−1\frac{1}{3} = 3^{-1} and 81=3481 = 3^4. The equation becomes (3−1)x−2=34(3^{-1})^{x-2} = 3^4. This simplifies to 3−1(x−2)=343^{-1(x-2)} = 3^4, or 3−x+2=343^{-x+2} = 3^4. Equating the exponents gives −x+2=4-x+2 = 4. Subtracting 2 from both sides gives −x=2-x = 2, so x=−2x = -2.