# 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 $\boxed{\textbf{(A)}}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions
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