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^{3 - 8 x} > 1

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

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

x (2^{x} - 1) (3^{x} - 9) (x - 3) < 0

\frac{( ( 3 ^{x} - 1 ) \cdot ( x - 1 ) ^{3} )}{( x \cdot ( x - 2 ) ^{5} )} \leq 0

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