2007 AMC 10A Problem 9

The bases are different ($3$ and $81$, then $125$ and $5$), but each pair comes from the same prime. Make the bases match, then compare the exponents.

1. First equation. $81 = 3^4$, so $81^{b+2} = (3^4)^{b+2} = 3^{4(b+2)}$. The equation becomes $3^a = 3^{4b+8}$. Same base, so the exponents are equal:
$$a = 4b + 8.$$
2. Second equation. $125 = 5^3$, so $125^b = 5^{3b}$. The equation becomes $5^{3b} = 5^{a-3}$, so
$$3b = a - 3.$$
3. Solve. Put $a = 4b + 8$ into the second equation:
$$3b = 4b + 8 - 3 = 4b + 5.$$
Take $4b$ from both sides: $-b = 5$, so $b = -5$.
4. Find $a$. $a = 4(-5) + 8 = -20 + 8 = -12$.
5. Check the second equation: $3(-5) = -15$ and $-12 - 3 = -15$. Correct.
6. Multiply. $ab = (-12)(-5) = 60$.

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

If $3^x = 3^y$, then $x = y$: different exponents always give different powers. So once both sides use the same base, an equation with powers turns into an ordinary equation. The rest is careful work with brackets and minus signs.

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 $\boxed{\textbf{(E)}\ 60}$.