Difference between revisions of "1950 AHSME Problems/Problem 26"
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<math> \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} </math> | <math> \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} </math> | ||
− | ==Solution== | + | ==Solution 1== |
We have <math>b=\log_{10}{10^b}</math>. Substituting, we find <math>\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}</math>. Using <math>\log{a}-\log{b}=\log{\dfrac{a}{b}}</math>, the left side becomes <math>\log_{10}{\dfrac{10^b}{n}}</math>. Because <math>\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}</math>, <math>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</math>. | We have <math>b=\log_{10}{10^b}</math>. Substituting, we find <math>\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}</math>. Using <math>\log{a}-\log{b}=\log{\dfrac{a}{b}}</math>, the left side becomes <math>\log_{10}{\dfrac{10^b}{n}}</math>. Because <math>\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}</math>, <math>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | adding <math>\log_{10} n</math> to both sides: | ||
+ | <cmath>\log_{10} m + \log_{10} n=b</cmath> | ||
+ | using the logarithm property <cmath>\log_a {b} + \log_a {c}=\log_a{bc}</cmath>: | ||
+ | <cmath>\log_{10} {mn}=b</cmath> | ||
+ | rewriting in exponential notation: | ||
+ | <cmath>10^b=mn</cmath> | ||
+ | <cmath>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 00:22, 10 June 2020
Contents
[hide]Problem
If , then
Solution 1
We have . Substituting, we find . Using , the left side becomes . Because , .
Solution 2
adding to both sides: using the logarithm property : rewriting in exponential notation:
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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