Difference between revisions of "2007 AMC 10A Problems/Problem 9"

Revision as of 10:50, 4 July 2013

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

$\text{(A)}\ -60 \qquad \text{(B)}\ -17 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 60$

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

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 $ab = 60\ \mathrm{(E)}$ .

2007 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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