1950 AHSME Problems/Problem 26

Problem

If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$

$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$

Solution

We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ , $m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}$ .

1950 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1950 AHSME Problems/Problem 26

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