What is log_{5}(2x-1)<log_{5}(x+7) ?

The solution to log_{5}(2x-1)<log_{5}(x+7) is 1/2 <x<8

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log_{5}(2x-1)<log_{5}(x+7)

lo g_{5} (2 x - 1) < lo g_{5} (x + 7)

\frac{1}{2} < x < 8

Interval Notation

(\frac{1}{2}, 8)

Decimal Notation

0.5 < x < 8

Solution steps

lo g_{5} (2 x - 1) < lo g_{5} (x + 7)

Apply log rules

2 x - 1 < x + 7 and 2 x - 1 > 0

2 x - 1 < x + 7 and 2 x - 1 > 0 : \frac{1}{2} < x < 8

\frac{1}{2} < x < 8

x \cdot ln (x^{3}) > 0

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

k \geq \frac{( lo g _{10} ( n ) )}{( lo g _{10} ( 2 ) )}

lo g_{x + 4} (5 x + 20) \leq lo g_{x + 4} ((x + 4)^{2})

lo g_{x} (x^{2} - 4 x + 4) \geq 0

What is log_{5}(2x-1)<log_{5}(x+7) ?
The solution to log_{5}(2x-1)<log_{5}(x+7) is 1/2 <x<8