For how many positive integers x is log_10(x - 40) + log_10(60 - x) < 2?

2014 AMC 12B Problem 20

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20.

For how many positive integers

x

\log_{10}(x - 40)

+ \log_{10}(60 - x) \lt 2?

10

18

19

20

infinitely many

Both logarithms require

40 \lt x \lt 60

Combining gives

(x-40)(60-x) \lt 100,

which is

(x-50)^2 \gt 0

The logarithms are defined only when

x - 40 \gt 0

and

60 - x \gt 0,

40 \lt x \lt 60.

Within this range the inequality becomes

(x-40)(60-x) \lt 100,

which expands to

x^2 - 100x + 2500 \gt 0,

i.e.

(x-50)^2 \gt 0.

This holds for every

x \ne 50.

The integers strictly between

40

and

60

except

50

are

41, \ldots, 49

and

51, \ldots, 59,

which is

18

values.

Thus, the correct answer is B.