What is ([2^{x-1}-1])/([2^{x+1)+1]}<2 ?

The solution to ([2^{x-1}-1])/([2^{x+1)+1]}<2 is True for all x

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([2^{x-1}-1])/([2^{x+1)+1]}<2

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

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

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

Apply exponent rules

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

Let

u = 2^{x}

\frac{[ \frac{u}{2} - 1 ]}{[ u \cdot 2 + 1 ]} < 2

\frac{[ \frac{u}{2} - 1 ]}{[ u \cdot 2 + 1 ]} < 2 : u < - \frac{6}{7} or u > - \frac{1}{2}

u < - \frac{6}{7} or u > - \frac{1}{2}

Substitute back

u = 2^{x}

2^{x} < - \frac{6}{7} or 2^{x} > - \frac{1}{2}

2^{x} < - \frac{6}{7} :

No Solution

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

True for all

x

Combine the intervals

No Solution or True for all x

Merge Overlapping Intervals

True for all x

2^{- x} < 0.25

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

4 \cdot 3^{x} < 3^{x} + 6

3^{x} > 4

3^{x} > 2

What is ([2^{x-1}-1])/([2^{x+1)+1]}<2 ?
The solution to ([2^{x-1}-1])/([2^{x+1)+1]}<2 is True for all x