2008 AMC 12B Problem 14 — Solution | LIVE by Po-Shen Loh

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2008 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2008 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12B solutions, or check the answer key.

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Concepts:logarithm circumference

Difficulty rating: 1630

14.

A circle has a radius of $\log_{10}(a^2)$ and a circumference of $\log_{10}(b^4).$ What is $\log_a b?$

$\dfrac{1}{4\pi}$

$\dfrac{1}{\pi}$

$\pi$

$2\pi$

$10^{2\pi}$

Solution:

The circumference is $2\pi$ times the radius, so $\log_{10}(b^4) = 2\pi \log_{10}(a^2).$

Rewriting, $4\log_{10} b = 4\pi \log_{10} a,$ hence $\log_{10} b = \pi \log_{10} a.$

Therefore $\log_a b = \dfrac{\log_{10} b}{\log_{10} a} = \pi.$

Thus, the correct answer is C.

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