What is log_{0.1}(4x^2-1)>log_{0.1}(3x) ?

The solution to log_{0.1}(4x^2-1)>log_{0.1}(3x) is 1/2 <x<1

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log_{0.1}(4x^2-1)>log_{0.1}(3x)

lo g_{0.1} (4 x^{2} - 1) > lo g_{0.1} (3 x)

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

Interval Notation

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

Decimal Notation

0.5 < x < 1

Solution steps

lo g_{0.1} (4 x^{2} - 1) > lo g_{0.1} (3 x)

Apply log rules

4 x^{2} - 1 < 3 x and 4 x^{2} - 1 > 0

4 x^{2} - 1 < 3 x and 4 x^{2} - 1 > 0 : \frac{1}{2} < x < 1

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

lo g_{2} (6 x + 2) > 0

0. 8^{l o g_{125} (x)} \geq 0. 8^{l o g_{10} (125)}

ln (x - e) > 0

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

lo g_{3} (2 x + 5) < 2

What is log_{0.1}(4x^2-1)>log_{0.1}(3x) ?
The solution to log_{0.1}(4x^2-1)>log_{0.1}(3x) is 1/2 <x<1