What is ((2^{1-x}-2^x+1))/((2^x-1))<= 0 ?

The solution to ((2^{1-x}-2^x+1))/((2^x-1)) = 1

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((2^{1-x}-2^x+1))/((2^x-1))<= 0

\frac{( 2 ^{1 - x} - 2 ^{x} + 1 )}{( 2 ^{x} - 1 )} \leq 0

x < 0 or x \geq 1

Interval Notation

(- \infty, 0) \cup [1, \infty)

Solution steps

\frac{( 2 ^{1 - x} - 2 ^{x} + 1 )}{( 2 ^{x} - 1 )} \leq 0

Identify the intervals

x < 0 or x \geq 1

2^{x} + 2^{- x + 1} - 3 < 0

(\frac{3}{2})^{1 + x^{2}} > (\frac{2}{3})^{3 x - 1}

\frac{( 2 ^{x - 1} - 1 )}{( 2 ^{x + 1} + 1 )} < 2

2^{3 - 8 x} > 1

(2^{x} + 1) (3^{x} + 1) (4^{x} + 1) < 0

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