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

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

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

inverse

f (x) = \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 )}

ln (x) + 5

f (x) = \frac{3}{1 - x}

f (x) = 6 x + 8 (x + 2)

f (x) = \frac{2}{x ^{3} + 27}

2 π r^{2} + 16 π r

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