2010 AMC 12B Problem 12

Problem 12 of 25IntermediateAlgebra

For what value of

x

does

\begin{aligned} &\log_{\sqrt2}\sqrt{x}+\log_2 x \\ &\quad {}+\log_4\left(x^2\right)+\log_8\left(x^3\right) \\ &\quad {}+\log_{16}\left(x^4\right)=40? \end{aligned}

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Solution

Let

L=\log_2 x.

Converting each term to base

2:

\log_{\sqrt2}\sqrt{x}=\dfrac{\frac{L}{2}}{\frac{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,

L=8

and

x=2^8=256.

Thus, the correct answer is D.