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

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2010 AMC 12A Problem 11

Below is the professionally curated solution for Problem 11 of the 2010 AMC 12A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:logarithm exponent

Difficulty rating: 1510

11.

The solution of the equation $7^{x+7}=8^x$ can be expressed in the form $x=\log_b 7^7.$ What is $b?$

$\dfrac{7}{15}$

$\dfrac{7}{8}$

$\dfrac{8}{7}$

$\dfrac{15}{8}$

$\dfrac{15}{7}$

Solution:

Since $x=\log_b 7^7,$ we have $b^x=7^7.$

Then $(7b)^x=7^x\cdot b^x=7^x\cdot7^7=7^{x+7}=8^x.$

Because $x\gt0,$ it follows that $7b=8,$ so $b=\dfrac{8}{7}.$

Thus, C is the correct answer.

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