2014 AMC 12B Problems/Problem 20

Problem

For how many positive integers $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$ ?

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

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.

Video Solution

https://youtu.be/RdIIEhsbZKw?t=1088

~ pi_is_3.14

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2014 AMC 12B Problems/Problem 20

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Solution

Video Solution

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