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}

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)