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Difference between revisions of "1950 AHSME Problems/Problem 26"

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<cmath>10^b=mn</cmath>
 
<cmath>10^b=mn</cmath>
 
<cmath>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</cmath>
 
<cmath>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</cmath>
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~Vndom
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==Solution 3==
 +
More simply, we can just simulate the problem, if we have <math>m = 10</math>, that means the right side must be 1, so the only way we can achieve that with distinct <math>n</math>, is if <math>b = 3</math>, and <math>n = 100</math>. With this we can look through the different answer choices substituting in <math>b</math>, <math>n</math>, and <math>m</math>, and find that <math>\boxed{\mathrm{(E)}}</math> is the only one that satisfies the question.
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~Shadow-18
  
 
==See Also==
 
==See Also==

Latest revision as of 18:38, 29 October 2020

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

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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All AHSME Problems and Solutions

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