What is log_{5}(2x-1)>log_{5}(4x) ?

The solution to log_{5}(2x-1)>log_{5}(4x) is No Solution

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log_{5}(2x-1)>log_{5}(4x)

lo g_{5} (2 x - 1) > lo g_{5} (4 x)

No Solution

Solution steps

lo g_{5} (2 x - 1) > lo g_{5} (4 x)

Apply log rules

2 x - 1 > 4 x and 4 x > 0

2 x - 1 > 4 x and 4 x > 0 :

No Solution

No Solution for x \in R

16 x^{2} - 9

6 - (- 8)

\frac{3}{8} + \frac{2}{3}

28 - \frac{x}{5} = \frac{x}{2}

x^{5} - 7 x^{4} + 12 x^{3}

What is log_{5}(2x-1)>log_{5}(4x) ?
The solution to log_{5}(2x-1)>log_{5}(4x) is No Solution