2014 AMC 12B Problems/Problem 20: Difference between revisions
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==Problem== | |||
For how many positive integers <math>x</math> is <math>\log_{10}(x-40) + \log_{10}(60-x) < 2</math> ? | |||
<math>\textbf{(A) }10\qquad | |||
\textbf{(B) }18\qquad | |||
\textbf{(C) }19\qquad | |||
\textbf{(D) }20\qquad | |||
\textbf{(E) }</math> infinitely many<math>\qquad</math> | |||
==Solution== | |||
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> | ||
| Line 4: | Line 15: | ||
<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 <math>\boxed{\textbf{(B)} 18}</math> integers. | ||
Revision as of 18:46, 20 February 2014
Problem
For how many positive integers 
is 
?
infinitely many
Solution
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 
integers.
