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

The inverse of f(x)=ln((x-1)/(x+1)) is (-1-e^x)/(e^x-1)

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inverse of f(x)=ln((x-1)/(x+1))

inverse

f (x) = ln (\frac{x - 1}{x + 1})

\frac{- 1 - e ^{x}}{e ^{x} - 1}

Solution steps

y = ln (\frac{x - 1}{x + 1})

Replace

x

with

y

x = ln (\frac{y - 1}{y + 1})

Solve for

y, x = ln (\frac{y - 1}{y + 1})

\frac{- 1 - e ^{x}}{e ^{x} - 1}

f (x) = \frac{- 3 x + 4}{- 2 x + 9}

f (x) = \frac{x - 10}{3}

f (x) = 9 x + 2

f (x) = \frac{4 - x}{2}

ln (1 + x^{2})

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