1954 AHSME Problems/Problem 38

Problem

If $\log 2 = .3010$ and $\log 3 = .4771$ , the value of $x$ when $3^{x+3} = 135$ is approximately

$\textbf{(A) \ } 5 \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$

Solution

Taking the logarithm in base $3$ of both sides, we get $x+3 = \log_3 135$ . Using the property $\log ab = \log a + \log b$ , we get $x+3 = \log_3 5 + \log_3 3^3$ , or $x = \log_3 5$ . Converting into base $10$ gives $x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}$ . Now, plugging in the values yeilds $\boxed{\textbf{(B) \ } 1.47 }$ .

1954 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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1954 AHSME Problems/Problem 38

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