Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.0 Junior Find the answer

If $\log_{10}2=a$ and $\log_{10}3=b$ , then $\log_{5}12=$ ?
$\textbf{(A)}\ \frac{a+b}{a+1}$ $\textbf{(B)}\ \frac{2a+b}{a+1}$ $\textbf{(C)}\ \frac{a+2b}{1+a}$ $\textbf{(D)}\ \frac{2a+b}{1-a}$ $\textbf{(E)}\ \frac{a+2b}{1-a}$

Your answer

Multiple choice: answer with the letter of the option you want.

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 $\boxed{\textbf{(D)}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.