# 2010 AMC 12B Problems/Problem 12

## Problem

For what value of $x$ does $$\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?$$ $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024$

## Solution 1 $$\log_{\sqrt{2}}\sqrt{x} + \log_2x + \log_4(x^2) + \log_8(x^3) + \log_{16}(x^4) = 40$$ $$\frac{1}{2} \frac{\log_2x}{\log_2\sqrt{2}} + \log_2x + \frac{2\log_2x}{\log_24} + \frac{3\log_2x}{\log_28} + \frac{4\log_2x}{\log_216} = 40$$ $$\log_2x + \log_2x + \log_2x + \log_2x + \log_2x = 40$$ $$5\log_2x = 40$$ $$\log_2x = 8$$ $$x = 256 \;\; (D)$$

## Solution 2

Using the fact that $\log_{x^n}{y^n} = \log_{x}{y}$, we see that the equation becomes $\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40$. Thus, $5\log_{2}{x} = 40$ and $\log_{2}{x} = 8$, so $x = 2^8 = 256$, or $\boxed{(D)}$.

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