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

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

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

inverse

y = \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 )}

f (x) = 34 e^{- 0.00625 x}

f (x) = 0.03 x + 0.63

(4 x - 1)^{2}

g (x) = 10 x - 4

f (x) = \frac{( 2 x - 7 )}{3 x}

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