# 1951 AHSME Problems/Problem 45

## Problem 45

If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$, then the only logarithm that cannot be found without the use of tables is:

$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$

## Solution

While $\log 17 = \log(8 + 9)$, we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore $\boxed{\textbf{(A)}\ \log 17}$.

As for the rest of the cases: $$\log\frac{5}{4} = \log\frac{10}{8} = \log 10 - \log 8 = 1 - \log 8$$ can be found; $$\log 15 = \log 3 + \log 5 = \frac{1}{2}\log 3^2 + \log\frac{10}{2} = \frac{1}{2} \log 9 + \log 10 - \log 2 = \frac{1}{2} \log 9 + 1 - \frac{1}{3} \log 8$$ can be found; $$\log 600 = \log 100 + \log 6 = 2 + \log 2 + \log 3 = 2 + \frac{1}{3} \log 8 + \frac{1}{2} \log 9$$ can be found; and $$\log .4 = \log\frac{4}{10} = \log 4 - \log 10 = 2 \log 2 - 1 = \frac{2}{3} \log 8 - 1$$ can be found.