# 1960 AHSME Problems/Problem 24

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## Problem

If $\log_{2x}216 = x$, where $x$ is real, then $x$ is:

$\textbf{(A)}\ \text{A non-square, non-cube integer}\qquad$ $\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad$ $\textbf{(C)}\ \text{An irrational number}\qquad$ $\textbf{(D)}\ \text{A perfect square}\qquad$ $\textbf{(E)}\ \text{A perfect cube}$

## Solution

Rewrite the equation to $(2x)^x = 216$. Since $2x$ is the base of the logarithm, we only need to check x-values that are positive.

With trial and error, $3$ is a solution because $(2 \cdot 3)^3 = 6^3 = 216. Since$(2x)^x$gets larger as x gets larger,$3$is the only solution, so the answer is$\boxed{\textbf{(A)}}\$.