Difference between revisions of "1950 AHSME Problems/Problem 26"
Ishankhare (talk | contribs) (→Solution) |
|||
| Line 8: | Line 8: | ||
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>. | ||
| + | |||
==See Also== | ==See Also== | ||
Revision as of 17:11, 27 March 2016
Problem
If 
, then 
Solution
We have 
. Substituting, we find 
. Using 
, the left side becomes 
. Because 
, 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
