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 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 $\boxed{\textbf{(C) }\frac{10}{7}}$.