What is log_{2}(x)+log_{2}(x-1)<1 ?

The solution to log_{2}(x)+log_{2}(x-1)<1 is 1<x<2

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log_{2}(x)+log_{2}(x-1)<1

lo g_{2} (x) + lo g_{2} (x - 1) < 1

1 < x < 2

Interval Notation

(1, 2)

Solution steps

lo g_{2} (x) + lo g_{2} (x - 1) < 1

Apply log rules

x (x - 1) < 2 and x (x - 1) > 0

x (x - 1) < 2 and x (x - 1) > 0 : - 1 < x < 0 or 1 < x < 2

- 1 < x < 0 or 1 < x < 2

Find positive values for logs:

x > 1

Combine the intervals

(- 1 < x < 0 or 1 < x < 2) and x > 1

Merge Overlapping Intervals

1 < x < 2

125 e^{0.7}

\frac{8}{( 2 + 2 ) ^{3}}

x^{2} + 3 x - 28 < 0

25 x^{2} + 30 x + 8 = 0

3 n^{2} - 8 n - 16

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