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 $8^{\frac{1}{3}} = 2$ , 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 $\fbox{B}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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1970 AHSME Problems/Problem 8

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