For what value of x does \log_\sqrt2√x+\log_2 x+\log_4(x^2)+\log_8(x^3)+\log_16(x^4)=40?

2010 AMC 12B Problem 12 — Solution

2010 AMC 12B Problem 12

Below is the professionally curated solution for Problem 12 of the 2010 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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Concepts:logarithm

Difficulty rating: 1500

12.

For what value of $x$ does $\log_{\sqrt2}\sqrt{x}+\log_2 x+\log_4\left(x^2\right)+\log_8\left(x^3\right)+\log_{16}\left(x^4\right)=40?$

$8$

$16$

$32$

$256$

$1024$

Solution:

Let $L=\log_2 x.$ Converting each term to base $2:$

$\log_{\sqrt2}\sqrt{x}=\dfrac{L/2}{1/2}=L,$ $\log_2 x=L,$ $\log_4 x^2=\dfrac{2L}{2}=L,$ $\log_8 x^3=\dfrac{3L}{3}=L,$ and $\log_{16}x^4=\dfrac{4L}{4}=L.$

The equation becomes $5L=40,$ so $L=8$ and $x=2^8=256.$

Thus, the correct answer is D.

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