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2007 AMC 10A Problems/Problem 9

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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

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

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

See also

2007 AMC 10A (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_10A_Problems/Problem_9"

Category: Introductory Algebra Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

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