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 $10^{-y} = \boxed{\textbf{(E)} \quad \frac{1}{5}}.$

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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
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1956 AHSME Problems/Problem 18

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