What is log_{10}(x)+log_{10}(x+3)<1 ?

The solution to log_{10}(x)+log_{10}(x+3)<1 is 0<x<2

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log_{10}(x)+log_{10}(x+3)<1

lo g_{10} (x) + lo g_{10} (x + 3) < 1

0 < x < 2

Interval Notation

(0, 2)

Solution steps

lo g_{10} (x) + lo g_{10} (x + 3) < 1

Apply log rules

x (x + 3) < 10 and x (x + 3) > 0

x (x + 3) < 10 and x (x + 3) > 0 : - 5 < x < - 3 or 0 < x < 2

- 5 < x < - 3 or 0 < x < 2

Find positive values for logs:

x > 0

Combine the intervals

(- 5 < x < - 3 or 0 < x < 2) and x > 0

Merge Overlapping Intervals

0 < x < 2

lo g_{x} (4 x - 3) \geq 2

lo g_{2} (lo g_{4} (lo g_{5} (x))) > 0

lo g_{3} (3 x - 4) > - 2

1 - 4 ln (x) < 17

lo g_{4} (x) < 0

What is log_{10}(x)+log_{10}(x+3)<1 ?
The solution to log_{10}(x)+log_{10}(x+3)<1 is 0<x<2