# 2007 AMC 10A Problems/Problem 9

## Problem

Real numbers $a$ and $b$ satisfy the equations $3^{a} = 81^{b + 2}$ and $125^{b} = 5^{a - 3}$. What is $ab$? $\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60$

## Solution 1 $$81^{b+2} = 3^{4(b+2)} = 3^a \Longrightarrow a = 4b+8$$

And $$125^{b} = 5^{3b} = 5^{a-3} \Longrightarrow a - 3 = 3b$$

Substitution gives $4b+8 - 3 = 3b \Longrightarrow b = -5$, and solving for $a$ yields $-12$. Thus $ab = 60\ \mathrm{(E)}$.

## Solution 1 another similar way

Simplify equation $1$, which is $3^a=81^{b+2}$, to $3^a=3^{4b+8}$.

And

Simplify equation $2$, which is $125^b=5^{a-3}$, to $5^{3b}=5^{a-3}$.

Now, eliminate the bases from the simplified equations $1$ and $2$ to arrive at $a=4b+8$ and $3b=a-3$. Rewrite equation $2$ so that it is in terms of $a$. That would be $a=3b+3$.

Since both equations are equal to $a$, and the values for $a$ and $b$ are constant for both equations, set the equations equal to each other. $4b+8=3b+3 \Longrightarrow b=-5$

Now plug $b$, which is $-5$, back into one of the two earlier equations. $4(-5)+8=a \Longrightarrow -20+8=a \Longrightarrow a=-12$ $(-12)(-5)=60$

Therefore the correct answer is $\mathrm{(E)}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 