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

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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