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

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

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

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

x > 2

Interval Notation

(2, \infty)

Solution steps

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

Apply log rules

(x - 1) x > 2

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

x < - 1 or x > 2

Find positive values for logs:

x > 1

Combine the intervals

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

Merge Overlapping Intervals

x > 2

\frac{1}{2 lo g _{10} ( n ^{2} - 1 )} > 1

lo g_{\frac{1}{3}} (3 x + 4) > lo g_{\frac{1}{3}} (x^{2} + 2)

lo g_{- 6 x - 5 x^{2}} (6^{x}) > 0

ln (x + 1) < x

lo g_{10} (x + 5) > 1

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