What is 1/(3^x+5)< 1/(3^{x+1)-1} ?

The solution to 1/(3^x+5)< 1/(3^{x+1)-1} is -1<x<1

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1/(3^x+5)< 1/(3^{x+1)-1}

\frac{1}{3 ^{x} + 5} < \frac{1}{3 ^{x + 1} - 1}

- 1 < x < 1

Interval Notation

(- 1, 1)

Solution steps

\frac{1}{3 ^{x} + 5} < \frac{1}{3 ^{x + 1} - 1}

Apply exponent rules

\frac{1}{3 ^{x} + 5} < \frac{1}{3 ^{x} \cdot 3 - 1}

Let

u = 3^{x}

\frac{1}{u + 5} < \frac{1}{u \cdot 3 - 1}

\frac{1}{u + 5} < \frac{1}{u \cdot 3 - 1} : u < - 5 or \frac{1}{3} < u < 3

u < - 5 or \frac{1}{3} < u < 3

Substitute back

u = 3^{x}

3^{x} < - 5 or \frac{1}{3} < 3^{x} < 3

3^{x} < - 5 :

No Solution

\frac{1}{3} < 3^{x} < 3 : - 1 < x < 1

Combine the intervals

No Solution or - 1 < x < 1

Merge Overlapping Intervals

- 1 < x < 1

(2 t^{- 3})^{3} (0.4 r)^{2}

\frac{x ^{2} - 9}{x - 1} \leq 0

2 w^{2} - 3 w - 3 = 0

\frac{1}{15} + \frac{1}{3}

\frac{1}{2} \cdot \frac{6}{5}

What is 1/(3^x+5)< 1/(3^{x+1)-1} ?
The solution to 1/(3^x+5)< 1/(3^{x+1)-1} is -1<x<1