What is log_{5}(2x-1)+log_{5}(x+2)>2 ?

The solution to log_{5}(2x-1)+log_{5}(x+2)>2 is x>3

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log_{5}(2x-1)+log_{5}(x+2)>2

lo g_{5} (2 x - 1) + lo g_{5} (x + 2) > 2

x > 3

Interval Notation

(3, \infty)

Solution steps

lo g_{5} (2 x - 1) + lo g_{5} (x + 2) > 2

Apply log rules

(2 x - 1) (x + 2) > 25

(2 x - 1) (x + 2) > 25 : x < - \frac{9}{2} or x > 3

x < - \frac{9}{2} or x > 3

Find positive values for logs:

x > \frac{1}{2}

Combine the intervals

(x < - \frac{9}{2} or x > 3) and x > \frac{1}{2}

Merge Overlapping Intervals

x > 3

(b^{2})^{\frac{3}{4}}

2 x^{2} - 5 x = - 12

4 x - 9 = 7 x

9 (10 - x) = 63

\frac{4 m}{3 mn} + 2

What is log_{5}(2x-1)+log_{5}(x+2)>2 ?
The solution to log_{5}(2x-1)+log_{5}(x+2)>2 is x>3