1961 AHSME Problems/Problem 30

Problem

If $\log_{10}2=a$ and $\log_{10}3=b$, then $\log_{5}12=$?

$\textbf{(A)}\ \frac{a+b}{a+1}\qquad \textbf{(B)}\ \frac{2a+b}{a+1}\qquad \textbf{(C)}\ \frac{a+2b}{1+a}\qquad \textbf{(D)}\ \frac{2a+b}{1-a}\qquad \textbf{(E)}\ \frac{a+2b}{1-a}$

Solution

From the change of base formula, $\log_{5}12 = \frac{\log_{10}12}{\log_{10}5}$.

For the numerator, $2 \cdot 2 \cdot 3 = 12$, so $\log_{10}12 = \log_{10}2 + \log_{10}2 + \log_{10}3 = 2a+b$.

For the denominator, note that $\log_{10}10 = 1$. Thus, $\log_{10}5 = \log_{10}10 - \log_{10}2 = 1 - a$.

Thus, the fraction is $\frac{2a+b}{1-a}$, so the answer is $\boxed{\textbf{(D)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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