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 ]} < 0

x < 0 or x > 1

Interval Notation

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

Solution steps

\frac{[ 2 ^{1 - x} - 2 ^{x} + 1 ]}{[ 2 ^{x} - 1 ]} < 0

Apply exponent rules

\frac{[ \frac{2}{2 ^{x}} - 2 ^{x} + 1 ]}{[ 2 ^{x} - 1 ]} < 0

f (x) > 0

we can multiply or devide both sides of inequality by

f (x)

2^{x}

is greater than

0

for all

x

\frac{[ \frac{2}{2 ^{x}} - 2 ^{x} + 1 ]}{[ 2 ^{x} - 1 ]} \cdot 2^{x} < 0 \cdot 2^{x}

Simplify

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

Let

v = 2^{2 x + 1}

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

\frac{2 ^{x} ( 2 ^{1 - x} - 2 ^{x} + 1 )}{2 ^{x} - 1} < 0 : x < 0 or x > 1

x < 0 or x > 1

Substitute back

v = 2^{2 x + 1}

x < 0 or x > 1

e^{x - 1} > 0

e^{- x} < 1

(\frac{2}{3})^{- x^{2} - 2 x + 12} \geq (\frac{9}{4})^{\frac{( x - 6 )}{2}}

2^{n} > 2 n + 1, n \geq 3

2^{x + 1} > 3^{x + 2}

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