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Difference between revisions of "2014 AMC 12B Problems/Problem 20"
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==Problem==
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For how many positive integers <math>x</math> is <math>\log_{10}(x-40) + \log_{10}(60-x) < 2</math> ?
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<math>\textbf{(A) }10\qquad
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\textbf{(B) }18\qquad
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\textbf{(C) }19\qquad
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\textbf{(D) }20\qquad
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\textbf{(E) }</math> infinitely many<math>\qquad</math>
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==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>
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Hence, we have integers from 41 to 49 and 51 to 59. There are 18 integers.
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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 $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$ ?

$\textbf{(A) }10\qquad \textbf{(B) }18\qquad \textbf{(C) }19\qquad \textbf{(D) }20\qquad \textbf{(E) }$ infinitely many$\qquad$

Solution

The domain of the LHS implies that \[40<x<60\] Begin from the left hand side \[\log_{10}[(x-40)(60-x)]<2\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{\textbf{(B)} 18}$ integers.

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