# 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 Begin from the left hand side $\[\log_{10}[(x-40)(60-x)]<2$$ $$(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

~ pi_is_3.14

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 