# Difference between revisions of "1970 AHSME Problems/Problem 8"

## Problem

If $a=\log_8 225$ and $b=\log_2 15$, then $\text{(A) } a=b/2\quad \text{(B) } a=2b/3\quad \text{(C) } a=b\quad \text{(D) } b=a/2\quad \text{(E) } a=3b/2$

## Solution

The solutions imply that finding the ratio $\frac{a}{b}$ will solve the problem. We compute $\frac{a}{b}$, use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base: $\frac{\log_8 225}{\log_2 15}$ $\frac{\frac{\ln 225}{\ln 8}}{\frac{\ln 15}{\ln 2}}$ $\frac{\ln 225 \ln 2}{\ln 15 \ln 8}$ $\frac{\ln 225}{\ln 15} \cdot \frac{\ln 2}{\ln 8}$ $\ln_{15} 225 \cdot \ln_8 2$

Since $15^2 = 225$, the first logarithm is $2$. Since $8^{\frac{1}{3} = 2$ (Error compiling LaTeX. ! Missing } inserted.), the second logarithm is $\frac{1}{3}$.

Thus, we have $\frac{a}{b} = \frac{2}{3}$, or $a = \frac{2}{3}b$, which is option $\fbox{B}$.

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