What is log_{10}(2-x)>log_{10}(1+2x) ?

The solution to log_{10}(2-x)>log_{10}(1+2x) is -1/2 <x< 1/3

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log_{10}(2-x)>log_{10}(1+2x)

lo g_{10} (2 - x) > lo g_{10} (1 + 2 x)

- \frac{1}{2} < x < \frac{1}{3}

Interval Notation

(- \frac{1}{2}, \frac{1}{3})

Decimal Notation

- 0.5 < x < 0.33333 \dots

Solution steps

lo g_{10} (2 - x) > lo g_{10} (1 + 2 x)

Apply log rules

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

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

- \frac{1}{2} < x < \frac{1}{3}

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

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

What is log_{10}(2-x)>log_{10}(1+2x) ?
The solution to log_{10}(2-x)>log_{10}(1+2x) is -1/2 <x< 1/3