What is the inverse of (2^x-1)/(2^x+1) ?

The inverse of (2^x-1)/(2^x+1) is (ln(\frac{-1-x)/(x-1))}{ln(2)}

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inverse of (2^x-1)/(2^x+1)

inverse

\frac{2 ^{x} - 1}{2 ^{x} + 1}

\frac{ln ( \frac{- 1 - x}{x - 1} )}{ln ( 2 )}

Solution steps

y = \frac{2 ^{x} - 1}{2 ^{x} + 1}

Replace

x

with

y

x = \frac{2 ^{y} - 1}{2 ^{y} + 1}

Solve for

y, x = \frac{2 ^{y} - 1}{2 ^{y} + 1}

\frac{ln ( \frac{- 1 - x}{x - 1} )}{ln ( 2 )}

y = x^{2} + 9, x \leq 0

- x^{2} + 2 x + 5

f (x) = \frac{1}{2 x}, x \neq = 0

f (x) = 13 x + 1

f (x) = \frac{( ( x - 13 ) )}{3}

What is the inverse of (2^x-1)/(2^x+1) ?
The inverse of (2^x-1)/(2^x+1) is (ln(\frac{-1-x)/(x-1))}{ln(2)}