Difference between revisions of "2014 AMC 12B Problems/Problem 20"

Latest revision as of 03:44, 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 $40<x<60$ Since $\log(a)$ is defined only when $a>0$ . $\[\]$ 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

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
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

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

@@ Line 9: / Line 9: @@
 ==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> Since <math>\log(a)</math> is defined only when <math>a>0</math>.
+<cmath></cmath>
+Begin from the left-hand side
 <cmath>\log_{10}[(x-40)(60-x)]<2</cmath>
 <cmath>(x-40)(60-x)<100</cmath>
 <cmath>-x^2+100x-2500<0</cmath>
@@ Line 16: / Line 19: @@
 <cmath>x \not = 50</cmath>
 Hence, we have integers from 41 to 49 and 51 to 59. There are <math>\boxed{\textbf{(B)} 18}</math> integers.
+==Solution 2==
+This solution is similar to the first solution, but perhaps more clear.
+By the properties of logarithms, <math>\log_{10}(x-40)+\log_{10}(60-x)=\log_{10}((x-40)(60-x))\implies\log_{10}((x-40)(60-x)).</math> Therefore, <math>(x-40)(60-x)<100.</math> 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 <math>x=50</math>, in which <math>(x-40)(60-x)=100.</math> This is the one point on the function that does not satisfy the given condition.
+We also know that, as noted in solution 1, <math>x</math> must be greater than <math>40</math> and less than <math>60</math>, since otherwise, <math>\log_{10}(x-40)</math> or <math>\log_{10}(60-x)</math> would be undefined. Thus, the possible values of <math>x</math> are the positive integers from <math>41</math> to <math>59,</math> inclusive, excluding <math>50.</math> Thus, there are <math>19-1=18</math> total integers.
+~ Technodoggo
 == Video Solution by OmegaLearn ==

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