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2014 AMC 12B Problem 20

Problem 20 of 25HarderAlgebra

For how many positive integers xx is log10(x40)\log_{10}(x - 40) +log10(60x)<2?+ \log_{10}(60 - x) \lt 2?

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Solution

The logarithms are defined only when x40>0x - 40 \gt 0 and 60x>0,60 - x \gt 0, so 40<x<60.40 \lt x \lt 60. Within this range the inequality becomes (x40)(60x)<100,(x-40)(60-x) \lt 100, which expands to x2100x+2500>0,x^2 - 100x + 2500 \gt 0, i.e. (x50)2>0.(x-50)^2 \gt 0. This holds for every x50.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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Concepts: logarithm · inequality · quadratic

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