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 2

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 $a$ and $b$ are the same number for both problems, set the equations equal to each other. $4b+8=3b+3$ $b=-5$

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

$(-12)(-5)=60$

Therefore the correct answer is E

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 10 Problems and Solutions

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2007 AMC 10A Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See also