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=3^4(b+2), which equals 3^a=3^4b+8.

And

Simplify equation 2 which is 125^b=5^a-3, to 5^3(b)=5^a-3, which equals 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 C

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2007 AMC 10A Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

See also