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2014 AMC 12B Problem 20

Problem 20 of 25HarderAlgebra

For how many positive integers xx is log⁡10(x−40)\log_{10}(x - 40) +log⁡10(60−x)<2?+ \log_{10}(60 - x) \lt 2?

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Solution

The logarithms are defined only when x−40>0x - 40 \gt 0 and 60−x>0,60 - x \gt 0, so 40<x<60.40 \lt x \lt 60. Within this range the inequality becomes (x−40)(60−x)<100,(x-40)(60-x) \lt 100, which expands to x2−100x+2500>0,x^2 - 100x + 2500 \gt 0, i.e. (x−50)2>0.(x-50)^2 \gt 0. This holds for every x≠50.x \ne 50. The integers strictly between 4040 and 6060 except 5050 are 41,…,4941, \ldots, 49 and 51,…,59,51, \ldots, 59, which is 1818 values. Thus, the correct answer is B.
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Tagged: logarithm · inequality · quadratic

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