AMC 10 · 2007 · #9

Grade 8 arithmetic

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

exponentssystems-of-equationslinear-equations-two-var convert-to-algebra ↑ Prerequisites: exponents

📏 Medium solution 💡 2 insights

Problem

Real numbers a and b satisfy two equations: 3^a = 81^(b+2) and 125^b = 5^(a-3). What is the product ab?

Pick an answer.

(A)

-60

(B)

-17

(C)

(D)

(E)

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Each equation mixes two different-looking bases, so Tool #7 (Identify Subproblems) treats them one at a time. The key that unlocks both is Tool #5 (Look for a Pattern): 81 and 125 are secretly powers of 3 and 5, so every term can be written on a single base. Once the bases match, equal powers force equal exponents, and each equation collapses into a plain linear relation between a and b. Two linear relations make a system that pins down a and b, and then ab is one multiplication. Tool #3 (Eliminate Possibilities) offers a fast sign check at the end.

1STEP 1

Write every base as a small prime power

Different bases block any comparison — but 81 = 3⁴ and 125 = 5³, so each equation can sit on a single base.

81 = 3⁴, 125 = 5³

2STEP 2

Match exponents in the first equation

Swap 81 for 3⁴: the exponent becomes 4(b+2), and equal powers of 3 force equal exponents, so a = 4b + 8.

3^a = 3⁴(b+2) → a = 4b + 8

3STEP 3

Match exponents in the second equation

Swap 125 for 5³: the left exponent is 3b, so 3b = a - 3, that is a = 3b + 3.

5³b = 5^a-3 → 3b = a - 3 → a = 3b + 3

4STEP 4

Combine the two relations

Both formulas describe the same a, so set them equal: 4b + 8 = 3b + 3 — the a cancels itself out.

4b + 8 = 3b + 3

5STEP 5

Solve for b, then a

Subtract 3b, then 8: b = -5. Feed that into a = 3b + 3 to get a = -12.

b = -5, a = 3(-5) + 3 = -12

6STEP 6

Multiply to get ab

Multiply the two values: ab = (-12)(-5). Negative times negative is positive, so ab = 60 — choice (E).

ab = (-12)(-5) = 60 → (E)

Answer

Check the pair a = -12, b = -5 against both original equations. First: 4b + 8 = 4(-5) + 8 = -12 = a, so 3^a = 3⁻¹² and 81^(b+2) = 81⁻³ = 3⁻¹² — they agree. Second: 3b = -15 and a - 3 = -15, so 125^b = 5⁻¹⁵ = 5^(a-3) — they agree too. The product (-12)(-5) = 60 is positive, and a quick sign check rules out the negative choices (A) and (B) as well as the small positives (C) and (D), leaving (E).

💡Key takeaway

When an equation hides different bases, rewrite them as powers of the same small number — then equal powers mean equal exponents, and the hard exponent problem becomes easy algebra.

Write every base as a small prime power
Match exponents in the first equation
Match exponents in the second equation
Combine the two relations
Solve for b, then a
Multiply to get ab