# 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 1

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 $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$, $m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}$.

## Solution 2

adding $\log_{10} n$ to both sides: $$\log_{10} m + \log_{10} n=b$$ using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ $$\log_{10} {mn}=b$$ rewriting in exponential notation: $$10^b=mn$$ $$m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}$$ ~Vndom

## Solution 3

More simply, we can just simulate the problem, if we have $m = 10$, that means the right side must be 1, so the only way we can achieve that with distinct $n$, is if $b = 3$, and $n = 100$. With this we can look through the different answer choices substituting in $b$, $n$, and $m$, and find that $\boxed{\mathrm{(E)}}$ is the only one that satisfies the question.