What is log_{2}(x)+log_{2}(x-1)>log_{2}(x+3) ?

The solution to log_{2}(x)+log_{2}(x-1)>log_{2}(x+3) is x>3

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log_{2}(x)+log_{2}(x-1)>log_{2}(x+3)

lo g_{2} (x) + lo g_{2} (x - 1) > lo g_{2} (x + 3)

x > 3

Interval Notation

(3, \infty)

Solution steps

lo g_{2} (x) + lo g_{2} (x - 1) > lo g_{2} (x + 3)

Move

lo g_{2} (x + 3)

to the left side

lo g_{2} (x) + lo g_{2} (x - 1) - lo g_{2} (x + 3) > 0

Apply log rules

\frac{x ( x - 1 )}{x + 3} > 1

\frac{x ( x - 1 )}{x + 3} > 1 : - 3 < x < - 1 or x > 3

- 3 < x < - 1 or x > 3

Find positive values for logs:

x > 1

Combine the intervals

(- 3 < x < - 1 or x > 3) and x > 1

Merge Overlapping Intervals

x > 3

lo g_{3} (x) > - 9

lo g_{10} (x^{2} - 1) > 0

ln (x) \leq x - 1

\frac{( lo g _{2} ( x + 1 ) )}{( x - 1 )} > 0

lo g_{3} (x^{2} - 8 x) < 2

What is log_{2}(x)+log_{2}(x-1)>log_{2}(x+3) ?
The solution to log_{2}(x)+log_{2}(x-1)>log_{2}(x+3) is x>3