1979 AHSME Problems/Problem 18

Problem 18

To the nearest thousandth, $\log_{10}2$ is $.301$ and $\log_{10}3$ is $.477$ . Which of the following is the best approximation of $\log_5 10$ ?

$\textbf{(A) }\frac{8}{7}\qquad \textbf{(B) }\frac{9}{7}\qquad \textbf{(C) }\frac{10}{7}\qquad \textbf{(D) }\frac{11}{7}\qquad \textbf{(E) }\frac{12}{7}$

Solution

Solution by e_power_pi_times_i

Notice that $\log_5 10 = \log_5 2 + 1 = \frac{1}{\log_2 5} + 1$ . So we are trying to find $\log_2 5$ . Denote $\log_{10}2$ as $x$ . Then $\frac{1}{x} = \log_2 10 = \log_2 5 + 1$ . Therefore $\log_2 5 = \frac{1}{x}-1$ , and plugging this in gives $\log_5 10 = \frac{1}{\frac{1}{x}-1}+1 = \frac{x}{1-x}+1 = \frac{1}{1-x}$ . Since $x$ is around $\frac{3}{10}$ , we substitute and get $\boxed{\textbf{(C) }\frac{10}{7}}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1979 AHSME Problems/Problem 18

Problem 18

Solution

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