If $\log_{10}{m}= b-\log_{10}{n}$, then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$
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 $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$, $m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}$.