1974 AHSME Problems/Problem 18

Problem

If $\log_8{3}=p$ and $\log_3{5}=q$ , then, in terms of $p$ and $q$ , $\log_{10}{5}$ equals

$\mathrm{(A)\ } pq \qquad \mathrm{(B) \ }\frac{3p+q}{5} \qquad \mathrm{(C) \ } \frac{1+3pq}{p+q} \qquad \mathrm{(D) \ } \frac{3pq}{1+3pq} \qquad \mathrm{(E) \ }p^2+q^2$

Solution

Notice that $\frac{1}{p}=\log_{3}{8}$ , so it would probably be easier to work in base $3$ . From change of base, $\log_{10}{5}=\frac{\log_{3}{5}}{\log_{3}{10}}$ . We're given that $\log_{3}{5}=q$ , so now we just need to find $\log_{3}{10}$ .

We have $\frac{1}{p}=\log_{3}{8}=3\log_{3}{2}$ , so $\log_{3}{2}=\frac{1}{3p}$ . Also, $\log_{3}{10}=\log_{3}{5}+\log{3}{2}=q+\frac{1}{3p}$ . Therefore, $\log_{10}{5}=\frac{q}{q+\frac{1}{3p}}=\frac{3pq}{3pq+1}, \boxed{\text{D}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1974 AHSME Problems/Problem 18

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