# 1953 AHSME Problems/Problem 22

## Problem

The logarithm of $27\sqrt{9}\sqrt{9}$ to the base $3$ is: $\textbf{(A)}\ 8\frac{1}{2} \qquad \textbf{(B)}\ 4\frac{1}{6} \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \text{none of these}$

## Solution $27\sqrt{9}\sqrt{9}$ can be rewritten as $3^3\cdot 3^\frac{1}{2}\cdot 3^\frac{2}{3}$. Using exponent rules, this simplifies to $3^\frac{25}{6}$. The problem wants us to find $\log_3{27\sqrt{9}\sqrt{9}}$. We just found that this is equal to $\log_3{3^\frac{25}{6}}$. Using logarithm rules, this is equal to $\frac{25}{6}\log_3{3}$, which is simply $\frac{25}{6}$. The answer is $\boxed{\text{(B)}\ 4\frac{1}{6}}$.

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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