1950 AHSME Problems/Problem 26: Difference between revisions
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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
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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