Difference between revisions of "2014 AMC 12B Problems/Problem 20"
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+ | ==Problem== | ||
+ | For how many positive integers <math>x</math> is <math>\log_{10}(x-40) + \log_{10}(60-x) < 2</math> ? | ||
+ | |||
+ | <math>\textbf{(A) }10\qquad | ||
+ | \textbf{(B) }18\qquad | ||
+ | \textbf{(C) }19\qquad | ||
+ | \textbf{(D) }20\qquad | ||
+ | \textbf{(E) }\text{infinitely many}\qquad</math> | ||
+ | |||
+ | ==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> | ||
+ | <cmath>(x-40)(60-x)<100</cmath> | ||
<cmath>-x^2+100x-2500<0</cmath> | <cmath>-x^2+100x-2500<0</cmath> | ||
<cmath>(x-50)^2>0</cmath> | <cmath>(x-50)^2>0</cmath> | ||
<cmath>x \not = 50</cmath> | <cmath>x \not = 50</cmath> | ||
− | Hence, we have integers from 41 to 49 and 51 to 59. There are 18 integers. | + | 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 == | ||
+ | https://youtu.be/RdIIEhsbZKw?t=1088 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Latest revision as of 22:03, 14 September 2023
Problem
For how many positive integers is
?
Solution
The domain of the LHS implies that Begin from the left hand side
Hence, we have integers from 41 to 49 and 51 to 59. There are
integers.
Solution 2
This solution is similar to the first solution, but perhaps more clear.
By the properties of logarithms, Therefore,
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
, in which
This is the one point on the function that does not satisfy the given condition.
We also know that, as noted in solution 1, must be greater than
and less than
, since otherwise,
or
would be undefined. Thus, the possible values of
are the positive integers from
to
inclusive, excluding
Thus, there are
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.