2010 AMC 12A Problem 11

Attempt Problem 11 of the 2010 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.

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11.

The solution of the equation

7^{x+7}=8^x

can be expressed in the form

x=\log_b 7^7.

What is

b?

\dfrac{7}{15}

\dfrac{7}{8}

\dfrac{8}{7}

\dfrac{15}{8}

\dfrac{15}{7}

x=\log_b 7^7

means

b^x=7^7

Multiply to combine the bases:

(7b)^x=7^x\cdot b^x=7^{x+7}

Since

x=\log_b 7^7,

we have

b^x=7^7.

Then

\begin{aligned} (7b)^x=7^x\cdot b^x &= 7^x\cdot7^7 \\ &= 7^{x+7}=8^x. \end{aligned}

Because

x\gt0,

it follows that

7b=8,

b=\dfrac{8}{7}.

Thus, C is the correct answer.

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