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.

~Shadow-18

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

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also