1958 AHSME Problems/Problem 20

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Problem

If $4^x - 4^{x - 1} = 24$, then $(2x)^x$ equals:

$\textbf{(A)}\ 5\sqrt{5}\qquad  \textbf{(B)}\ \sqrt{5}\qquad  \textbf{(C)}\ 25\sqrt{5}\qquad  \textbf{(D)}\ 125\qquad  \textbf{(E)}\ 25$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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