1956 AHSME Problems/Problem 18

Problem

If $10^{2y} = 25$, then $10^{ - y}$ equals:

$\textbf{(A)}\ -\frac{1}{5}\qquad \textbf{(B)}\ \frac{1}{625}\qquad \textbf{(C)}\ \frac{1}{50}\qquad \textbf{(D)}\ \frac{1}{25}\qquad \textbf{(E)}\ \frac{1}{5}$


Solution

If $10^{2y}=(10^{y})^2 = 25$, then $10^y = 5$ and $10^{-y} = \boxed{\textbf{(E)} \quad \frac{1}{5}}.$


See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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