Difference between revisions of "2014 AMC 12B Problems/Problem 20"
(Created page with "The domain of the LHS implies that <cmath>40<x<60</cmath> Begin from the left hand side <cmath>log_{10}[(x-40)(60-x)]<2</cmath> <cmath>-x^2+100x-2500<0</cmath> <cmath>(x-50)^2>0<...") |
|||
| Line 1: | Line 1: | ||
The domain of the LHS implies that <cmath>40<x<60</cmath> Begin from the left hand side | The domain of the LHS implies that <cmath>40<x<60</cmath> Begin from the left hand side | ||
| − | <cmath>log_{10}[(x-40)(60-x)]<2</cmath> | + | <cmath>\log_{10}[(x-40)(60-x)]<2</cmath> |
<cmath>-x^2+100x-2500<0</cmath> | <cmath>-x^2+100x-2500<0</cmath> | ||
<cmath>(x-50)^2>0</cmath> | <cmath>(x-50)^2>0</cmath> | ||
<cmath>x \not = 50</cmath> | <cmath>x \not = 50</cmath> | ||
Hence, we have integers from 41 to 49 and 51 to 59. There are 18 integers. | Hence, we have integers from 41 to 49 and 51 to 59. There are 18 integers. | ||
Revision as of 16:48, 20 February 2014
The domain of the LHS implies that ![\[40<x<60\]](http://latex.artofproblemsolving.com/e/6/1/e61ac4e06141f931a998331029b57ad65f67c533.png)
Begin from the left hand side
![\[\log_{10}[(x-40)(60-x)]<2\]](http://latex.artofproblemsolving.com/6/7/4/6749fe9dcc8059ce47ccd96766d2b0d9d1f0eb3a.png)
![\[-x^2+100x-2500<0\]](http://latex.artofproblemsolving.com/8/c/b/8cb07d38ab8be9ccbb57dd02425ab79f4a9da74d.png)
![\[(x-50)^2>0\]](http://latex.artofproblemsolving.com/d/c/7/dc75fe198e27a34d5b4d4876eed002ce2eb24c26.png)
![\[x \not = 50\]](http://latex.artofproblemsolving.com/3/9/c/39c9aefab22bafbe506412f1eeed25a27f7cf9d5.png)
Hence, we have integers from 41 to 49 and 51 to 59. There are 18 integers.
