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

The solution to log_{sqrt(2)}(1-2x)<log_{sqrt(2)}(x+1) is 0<x< 1/2

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log_{sqrt(2)}(1-2x)<log_{sqrt(2)}(x+1)

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

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

Interval Notation

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

Decimal Notation

0 < x < 0.5

Solution steps

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

Apply log rules

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

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

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

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

lo g_{3} (x + 1) + lo g_{3} (x + 7) \geq 3

lo g_{x} (6 - x) > 2

lo g_{4} (- x) > 0

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

What is log_{sqrt(2)}(1-2x)<log_{sqrt(2)}(x+1) ?
The solution to log_{sqrt(2)}(1-2x)<log_{sqrt(2)}(x+1) is 0<x< 1/2