Difference between revisions of "1950 AHSME Problems/Problem 26"
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(→Solution 2) |
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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> | ||
| + | ~Vndom | ||
==See Also== | ==See Also== | ||
Revision as of 00:27, 10 June 2020
Contents
Problem
If 
, then 
Solution 1
We have 
. Substituting, we find 
. Using 
, the left side becomes 
. Because 
, 
.
Solution 2
adding 
to both sides:
![\[\log_{10} m + \log_{10} n=b\]](http://latex.artofproblemsolving.com/a/c/d/acd4aae819e142a2d1f955adf4173b699ab685c3.png)
using the logarithm property: 
![\[\log_{10} {mn}=b\]](http://latex.artofproblemsolving.com/c/b/d/cbded8a2340f38f2e9c93e2c89067f6457b1429b.png)
rewriting in exponential notation:
![\[10^b=mn\]](http://latex.artofproblemsolving.com/8/b/1/8b10c86f05dc25ebc5f3cea860f97588d503106d.png)
![\[m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}\]](http://latex.artofproblemsolving.com/9/9/d/99d0c6f5d34ea6d45d7ce6820dc2985563f111a6.png)
~Vndom
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. 
