2010 AMC 12A Problem 11 — Solution | LIVE by Po-Shen Loh

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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 $b?$

$\dfrac{7}{15}$

$\dfrac{7}{8}$

$\dfrac{8}{7}$

$\dfrac{15}{8}$

$\dfrac{15}{7}$

Answer: C

Concepts:logarithm exponent

Difficulty rating: 1510

Solution:

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 $7b=8,$ so $b=\dfrac{8}{7}.$

Thus, C is the correct answer.

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