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2007 AMC 10A Problem 9

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2007 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 2 of 5 · about 2 min · position P1-10

Real numbers $a$ and $b$ satisfy the equations $3^{a} = 81^{b + 2}$ and $125^{b} = 5^{a - 3}$ . What is $ab$ ?

A)
$-60$
B)
$-17$
C)
$9$
D)
$12$
E)
$60$

Answer

Key insight

Write 81 as 3^4 and 125 as 5^3 so each equation becomes an equality of exponents, a linear system in a and b.

Solution

Express everything with prime bases. Since $81 = 3^4$ , the first equation reads $3^a = 3^{4(b+2)}$ , so
$a = 4b + 8.$
Since $125 = 5^3$ , the second equation reads $5^{3b} = 5^{a-3}$ , so
$3b = a - 3.$
Substituting the first into the second: $3b = 4b + 8 - 3$ , giving $b = -5$ , and then $a = 4(-5) + 8 = -12$ .

Therefore $ab = (-12)(-5) = 60$ .

The answer is $\boxed{\textbf{(E)}\ 60}$ .

Why this works

Exponential equations with the same underlying prime base collapse to linear equations in the exponents, because $x \mapsto 3^x$ is one-to-one. The remaining work is a tiny linear system; the only care needed is bookkeeping with parentheses and signs.

The trap

Forgetting to distribute the 4 over b + 2 (writing 4b + 2), or losing a sign when both a and b turn out negative.

Common mistakes

Forgetting to distribute the $4$ over $b + 2$ (writing $4b + 2$ ), or losing a sign when both $a$ and $b$ turn out negative.
Answering $-60$ by mishandling the product of two negatives, or reporting $a + b = -17$ instead of $ab$ .

Techniques

Set up the equation/formula and compute; no special trick needed