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 )}

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

- \frac{1}{x} + 2

2 lo g_{10} (x + 3)

\frac{x ^{2} + 5 x + 6}{x + 2}

f (x) = (x + 5) (x - 5)

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)}