SAT Exponential Equations Practice Questions | Download PDF

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SAT exponential equation practice questions assess your ability to solve for a variable in an exponent, identify the appropriate exponential model for a real-world growth or decay scenario, and decipher the components of an exponential expression. There are 55 SAT-style exponential equation questions on this page, with full solutions, answer options, and trap notes.The set progresses from simple exponential solving to more complex multi-step reasoning, matching bases, growth/decay word problems, and model parameter interpretation.

In This Guide – 55 Exponential Equations Practice Questions

1. What does the SAT test in exponential equations?
2. How do you solve basic exponential equations?
3. How do you handle matching bases and rewriting exponents?
4. How are exponential equations used in growth & decay word problems?
5. How do you interpret parts of an exponential model?
6. What do hard exponential equation questions look like?
7. What mistakes cost students points?
8. How should you study exponential equations in 2 weeks?
9. Frequently asked questions

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SAT strategy: Prior to solving, find out if both sides can be rebuilt using the same basis. If so, make sure the exponents are equivalent. If not, determine if the question actually asks you to find a model parameter instead than solving for a number.

Solve for x: 2^x = 32

A) x = 5    B) x = 4    C) x = 6    D) x = 16

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32 = 2⁵, so 2^x = 2⁵ means x = 5. Answer: A

Solve for x: 3^x = 81

A) x = 4    B) x = 3    C) x = 27    D) x = 5

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81 = 3⁴, so x = 4. Answer: A

Solve for x: 5^(x+1) = 125

A) x = 2    B) x = 3    C) x = 1    D) x = 4

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125 = 5³, so x + 1 = 3, giving x = 2. Answer: A

Solve for x: 4^(2x) = 64

A) x = 1.5    B) x = 3    C) x = 2    D) x = 0.75

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64 = 4³, so 2x = 3, giving x = 1.5. Answer: A

Solve for x: 10^x = 1000

A) x = 3    B) x = 2    C) x = 4    D) x = 100

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1000 = 10³, so x = 3. Answer: A

Solve for x: 2^(x-3) = 16

A) x = 7    B) x = 4    C) x = 13    D) x = 1

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16 = 2⁴, so x − 3 = 4, giving x = 7. Answer: A

Solve for x: 6^x = 1

A) x = 0    B) x = 1    C) x = 6    D) No solution

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Any nonzero base raised to the 0 power equals 1, so x = 0. Answer: A

Solve for x: 3^(2x+1) = 27

A) x = 1    B) x = 2    C) x = 0.5    D) x = 1.5

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27 = 3³, so 2x + 1 = 3 → 2x = 2 → x = 1. Answer: A

Solve for x: 9^x = 27

A) x = 1.5    B) x = 3    C) x = 2    D) x = 0.5

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Rewrite with base 3: 9^x = (3²)^x = 3^(2x), and 27 = 3³. So 2x = 3, giving x = 1.5. Answer: A

Solve for x: 2^x = 1/8

A) x = -3    B) x = 3    C) x = -8    D) x = 1/3

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1/8 = 2⁻³, so x = -3. Answer: A

Solve for x: 4^x = 8^(x-1)

A) x = 3    B) x = 2    C) x = 6    D) x = 1

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Rewrite with base 2: 2^(2x) = 2^(3x-3). So 2x = 3x − 3 → -x = -3 → x = 3. Answer: A

Solve for x: 5 · 2^x = 40

A) x = 3    B) x = 4    C) x = 8    D) x = 2

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Divide both sides by 5: 2^x = 8 = 2³, so x = 3. Answer: A

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Solve for x: 27^x = 9

A) x = 2/3    B) x = 3/2    C) x = 1/3    D) x = 3

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Rewrite with base 3: (3³)^x = 3², so 3x = 2, giving x = 2/3. Answer: A

Solve for x: (1/2)^x = 8

A) x = -3    B) x = 3    C) x = -8    D) x = 1/8

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(1/2)^x = 2⁻ˣ, and 8 = 2³. So -x = 3, giving x = -3. Answer: A

Solve for x: √(2^x) = 16

A) x = 8    B) x = 4    C) x = 16    D) x = 32

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√(2^x) = 2^(x/2), and 16 = 2⁴. So x/2 = 4, giving x = 8. Answer: A

Solve for x: 16^x = 2

A) x = 1/4    B) x = 4    C) x = 1/2    D) x = 2

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(2⁴)^x = 2¹, so 4x = 1, giving x = 1/4. Answer: A

Solve for x: 25^(x-1) = 125^x

A) x = -2    B) x = 2    C) x = -1    D) x = 1

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Rewrite with base 5: 5^(2x-2) = 5^(3x). So 2x − 2 = 3x → -2 = x. Answer: A

Solve for x: 8^x = 1/4

A) x = -2/3    B) x = 2/3    C) x = -3/2    D) x = -4

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2^(3x) = 2⁻², so 3x = -2, giving x = -2/3. Answer: A

Solve for x: 3^(x²) = 3^(4x – 3)

A) x = 1 or x = 3    B) x = -1 or x = -3    C) x = 3 only    D) x = 4

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Since the bases match, set exponents equal: x² = 4x − 3 → x² − 4x + 3 = 0 → (x−1)(x−3) = 0, so x = 1 or x = 3. Answer: A

Solve for x: 4^x · 4^3 = 4^9

A) x = 6    B) x = 12    C) x = 3    D) x = 27

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4^(x+3) = 4⁹, so x + 3 = 9, giving x = 6. Answer: A

Solve for x: 2^(x+2) = 3 · 2^x

A) No solution    B) x = 1    C) x = 2    D) x = 3

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2^(x+2) = 2^x · 2² = 4 · 2^x. Setting 4 · 2^x = 3 · 2^x gives 4 = 3, which is never true, so there is no solution. Answer: A

Solve for x: 9^x = 3^(x+2)

A) x = 2    B) x = 1    C) x = 4    D) x = -2

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3^(2x) = 3^(x+2), so 2x = x + 2, giving x = 2. Answer: A

Solve for x: (1/4)^(x-1) = 64

A) x = -2    B) x = 2    C) x = -3    D) x = 4

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(1/4)^(x-1) = 2^(-2(x-1)) = 2^(-2x+2), and 64 = 2⁶. So -2x + 2 = 6 → -2x = 4 → x = -2. Answer: A

A bacterial population doubles per hour from a starting point of 200. After t hours, which equation represents the population P?

A) P = 200(2)^t    B) P = 200(t)^2    C) P = 2(200)^t    D) P = 200 + 2t

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Doubling every hour means the growth factor is 2 raised to the number of hours: P = 200(2)^t. Answer: A

An investment of $5,000 grows at 4% annually. Which equation models the value V after t years?

A) V = 5000(1.04)^t    B) V = 5000(0.04)^t    C) V = 5000(1.4)^t    D) V = 5000(4)^t

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A 4% annual growth rate gives a growth factor of 1 + 0.04 = 1.04: V = 5000(1.04)^t. Answer: A

A car worth $28,000 depreciates 12% per year. Which equation models its value V after t years?

A) V = 28000(0.88)^t    B) V = 28000(0.12)^t    C) V = 28000(1.12)^t    D) V = 28000 − 12t

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A 12% decrease gives a decay factor of 1 − 0.12 = 0.88: V = 28000(0.88)^t. Answer: A

The half-life of a radioactive sample is ten days. Which formula represents the mass m after t days if it begins at 80 grams?

A) m = 80(1/2)^(t/10)    B) m = 80(1/2)^(10t)    C) m = 80(2)^(t/10)    D) m = 80 − (t/10)

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Half-life problems use (1/2) raised to (time ÷ half-life period): m = 80(1/2)^(t/10). Answer: A

Using the sample from Question 27, how much mass remains after 30 days?

A) 10 grams    B) 20 grams    C) 40 grams    D) 8 grams

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m = 80(1/2)^(30/10) = 80(1/2)³ = 80(1/8) = 10 grams. Answer: A

A town’s population is modeled by P = 12000(1.03)^t. What is the annual percent growth rate?

A) 3%    B) 1.03%    C) 30%    D) 103%

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The growth factor 1.03 corresponds to 1 + 0.03, so the annual growth rate is 3%. Answer: A

A machine’s value is modeled by V = 15000(0.9)^t. What is the annual percent decrease?

A) 10%    B) 90%    C) 9%    D) 0.9%

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The decay factor 0.9 corresponds to 1 − 0.10, so the value decreases by 10% each year. Answer: A

A culture of bacteria triples every 4 hours, starting at 50. Which equation models the population N after t hours?

A) N = 50(3)^(t/4)    B) N = 50(3)^(4t)    C) N = 50(4)^(t/3)    D) N = 50(3t)^4

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Tripling every 4 hours means the exponent is time divided by the tripling period: N = 50(3)^(t/4). Answer: A

Using the model from Question 31, what is the bacteria population after 12 hours?

A) 1,350    B) 450    C) 4,050    D) 150

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N = 50(3)^(12/4) = 50(3)³ = 50(27) = 1,350. Answer: A

A savings account starts at $2,000 and earns 6% interest compounded annually. Approximately how much is in the account after 3 years?

A) $2,382    B) $2,360    C) $2,400    D) $2,120

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V = 2000(1.06)³ ≈ 2000(1.191) ≈ $2,382. Answer: A

A population P₀ decreases by 25% every 6 years. Which expression gives the decay factor per year?

A) 0.75^(1/6)    B) 0.75^6    C) 0.25^(1/6)    D) (1/6)^0.75

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If the 6-year factor is 0.75, the annual factor r satisfies r⁶ = 0.75, so r = 0.75^(1/6). Answer: A

A website’s daily visitors double every 5 days, starting at 300 visitors. Which equation gives the number of visitors V after t days?

A) V = 300(2)^(t/5)    B) V = 300(5)^(t/2)    C) V = 300(2)^(5t)    D) V = 300 + 2(t/5)

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Doubling every 5 days: V = 300(2)^(t/5). Answer: A

A drug concentration in the bloodstream decreases by half every 8 hours. If the initial dose creates a concentration of 60 mg/L, after how many hours will the concentration first drop below 10 mg/L?

A) 24 hours    B) 16 hours    C) 32 hours    D) 20 hours

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After 24 hours (3 half-lives): 60(1/2)³ = 7.5 mg/L, which is below 10. After 16 hours (2 half-lives): 60(1/2)² = 15 mg/L, still above 10. So the concentration first drops below 10 at 24 hours. Answer: A

A stock price P is modeled by P = 40(1.08)^t, where t is in years. What was the original stock price?

A) $40    B) $1.08    C) $43.20    D) $8

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At t = 0, P = 40(1.08)⁰ = 40(1) = $40. Answer: A

A forest’s tree count is modeled by N = 8000(0.95)^t. Approximately how many years until the population falls to about half its original size?

A) about 14 years    B) about 7 years    C) about 20 years    D) about 10 years

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Testing t = 14: 0.95^14 ≈ 0.488, close to half. Testing t = 7: 0.95^7 ≈ 0.698, not yet half. So about 14 years is the better estimate. Answer: A

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In the model P = 500(1.02)^t, where P is population and t is years, what does 500 represent?

A) The initial population    B) The annual growth rate    C) The population after 1 year    D) The number of years

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In y = a(b)^t form, a is the starting value, so 500 is the initial population at t = 0. Answer: A

In the model V = 3000(0.85)^t, where V is value in dollars and t is years, what does 0.85 represent?

A) The value decreases by 15% each year    B) The value decreases by 85% each year    C) The value increases by 85% each year    D) The initial value

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A decay factor of 0.85 corresponds to 1 − 0.15, meaning a 15% annual decrease. Answer: A

In the model N = 150(1.12)^t, what is the growth factor applied each year?

A) 1.12    B) 0.12    C) 12    D) 150

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The base of the exponential expression, 1.12, is the growth factor applied each time period. Answer: A

In the model A = 1200(1.005)^(12t), where t is years, what does the exponent 12t most likely represent?

A) The number of months, since interest compounds monthly    B) The number of years    C) The annual interest rate    D) The initial deposit

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Multiplying years by 12 converts to months, suggesting monthly compounding at a monthly rate of 0.5%. Answer: A

In the model M = 40(0.5)^(t/12), where M is mass in grams and t is in hours, what does 12 represent?

A) The half-life in hours    B) The initial mass    C) The number of half-lives    D) The decay rate as a percent

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Since the base is 0.5 (representing halving), the divisor in the exponent is the half-life: 12 hours. Answer: A

Two models describe population growth: P₁ = 400(1.03)^t and P₂ = 400(1.07)^t. Which statement is true?

A) P₂ grows faster than P₁    B) P₁ grows faster than P₂    C) They grow at the same rate    D) P₂ starts larger than P₁

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A larger growth factor (1.07 vs. 1.03) means faster growth, and both start at the same initial value of 400. Answer: A

A model gives C = C₀(1 − r)^t for a decaying quantity. If r = 0.2, what percent of the original quantity remains after each time period?

A) 80%    B) 20%    C) 60%    D) 120%

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1 − r = 1 − 0.2 = 0.8, meaning 80% remains after each period. Answer: A

In the model H = 90(0.75)^n, where H is the height in feet after n bounces of a ball, what does 90 represent?

A) The initial drop height    B) The height after 1 bounce    C) The percent height retained    D) The number of bounces

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At n = 0, H = 90(0.75)⁰ = 90, so 90 is the initial height before any bounces. Answer: A

In the model N = a(b)^t, if the quantity doubles every 3 years and starts at 50, what are the values of a and b?

A) a = 50, b = 2^(1/3)    B) a = 50, b = 2    C) a = 2, b = 50^(1/3)    D) a = 3, b = 50

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The initial value a = 50. Since the quantity doubles every 3 years, the annual growth factor satisfies b³ = 2, so b = 2^(1/3). Answer: A

If 2^x = 5, what is the value of 2^(x+2)?

A) 20    B) 10    C) 25    D) 7

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2^(x+2) = 2^x · 2² = 5 · 4 = 20. Answer: A

If 3^a = 7, what is the value of 9^a?

A) 49    B) 14    C) 21    D) 63

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9^a = (3²)^a = (3^a)² = 7² = 49. Answer: A

Two populations are modeled by A = 100(1.5)^t and B = 800(1.1)^t. Which statement about the populations is true for large values of t?

A) A will eventually exceed B, despite starting smaller    B) B will always be larger than A    C) A and B will always be equal    D) Neither model changes over time

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Even though A starts smaller, its larger growth factor (1.5 vs. 1.1) means it will eventually overtake B as t increases. Answer: A

If 4^x = 10, what is the value of 2^x?

A) √10    B) 5    C) 10²    D) 100

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4^x = (2²)^x = (2^x)² = 10, so 2^x = √10. Answer: A

If f(t) = a(b)^t models exponential growth, and f(0) = 20 and f(2) = 45, what is the value of b?

A) 1.5    B) 2.25    C) 1.25    D) 2

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f(0) = a = 20. Then f(2) = 20b² = 45 → b² = 2.25 → b = 1.5. Answer: A

If 5^x = 12, what is the value of 5^(2x)?

A) 144    B) 24    C) 60    D) 17

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5^(2x) = (5^x)² = 12² = 144. Answer: A

A quantity Q grows according to Q = 500(1.2)^t. At what value of t does Q first exceed 1,000?

A) between t = 3 and t = 4    B) between t = 1 and t = 2    C) between t = 5 and t = 6    D) exactly t = 4

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At t = 3: 500(1.2)³ ≈ 864. At t = 4: 500(1.2)⁴ ≈ 1037. So Q first exceeds 1,000 between t = 3 and t = 4. Answer: A

If h(x) = 2^(x) and k(x) = 2^(x-3), how does the graph of k compare to the graph of h?

A) k is h shifted right 3 units    B) k is h shifted left 3 units    C) k is h shifted up 3 units    D) k is h stretched vertically by 3

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Replacing x with (x − 3) in the exponent shifts the graph 3 units to the right. Answer: A

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