What is 1/(2^x-1)> 1/(1-2^{x-1)} ?

The solution to 1/(2^x-1)> 1/(1-2^{x-1)} is 0 1

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1/(2^x-1)> 1/(1-2^{x-1)}

\frac{1}{2 ^{x} - 1} > \frac{1}{1 - 2 ^{x - 1}}

0 < x < 2 - lo g_{2} (3) or x > 1

Interval Notation

(0, 2 - lo g_{2} (3)) \cup (1, \infty)

Decimal Notation

0 < x < 0.41503 \dots or x > 1

Solution steps

\frac{1}{2 ^{x} - 1} > \frac{1}{1 - 2 ^{x - 1}}

Apply exponent rules

\frac{1}{2 ^{x} - 1} > \frac{1}{1 - \frac{2 ^{x}}{2}}

Let

u = 2^{x}

\frac{1}{u - 1} > \frac{1}{1 - \frac{u}{2}}

\frac{1}{u - 1} > \frac{1}{1 - \frac{u}{2}} : 1 < u < \frac{4}{3} or u > 2

1 < u < \frac{4}{3} or u > 2

Substitute back

u = 2^{x}

1 < 2^{x} < \frac{4}{3} or 2^{x} > 2

1 < 2^{x} < \frac{4}{3} : 0 < x < 2 - lo g_{2} (3)

2^{x} > 2 : x > 1

Combine the intervals

0 < x < 2 - lo g_{2} (3) or x > 1

((\frac{3}{7})^{x})^{2} < (\frac{7}{3})^{4} x - 2

(\frac{2}{5})^{\frac{( 1 - 5 x )}{( 4 + 16 )}} < \frac{25}{4}

4^{x - 1} + 2^{x + 1} \geq 2^{5}

2^{x + 2} > (\frac{1}{4})^{\frac{1}{x}}

e^{- 2 x^{2}} - 1 > 0

What is 1/(2^x-1)> 1/(1-2^{x-1)} ?
The solution to 1/(2^x-1)> 1/(1-2^{x-1)} is 0<x<2-log_{2}(3)\lor x>1