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Difference between revisions of "2014 AMC 12B Problems/Problem 20"
m (→‎Solution: clarification about domain)
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==Solution==
 
==Solution==
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The domain of the LHS implies that <cmath>40<x<60</cmath> Begin from the left hand side
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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>
 +
Since <math>\log(a)</math> is defined only when <math>a>0</math>
 
<cmath>(x-40)(60-x)<100</cmath>
 
<cmath>(x-40)(60-x)<100</cmath>
 
<cmath>-x^2+100x-2500<0</cmath>
 
<cmath>-x^2+100x-2500<0</cmath>

Revision as of 03:42, 2 November 2024

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 \[\log_{10}[(x-40)(60-x)]<2\] Since $\log(a)$ is defined only when $a>0$ \[(x-40)(60-x)<100\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{\textbf{(B)} 18}$ integers.

Solution 2

This solution is similar to the first solution, but perhaps more clear.

By the properties of logarithms, $\log_{10}(x-40)+\log_{10}(60-x)=\log_{10}((x-40)(60-x))\implies\log_{10}((x-40)(60-x)).$ Therefore, $(x-40)(60-x)<100.$ We can see that this function is concave down (that is, it looks like an upside-down U shape); that is, it has a maximum value. This maximum value is attained at $x=50$, in which $(x-40)(60-x)=100.$ This is the one point on the function that does not satisfy the given condition.

We also know that, as noted in solution 1, $x$ must be greater than $40$ and less than $60$, since otherwise, $\log_{10}(x-40)$ or $\log_{10}(60-x)$ would be undefined. Thus, the possible values of $x$ are the positive integers from $41$ to $59,$ inclusive, excluding $50.$ Thus, there are $19-1=18$ total integers.

~ Technodoggo

Video Solution by OmegaLearn

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

~ pi_is_3.14

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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
All AMC 12 Problems and Solutions

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