What is the inverse of ((e^x+1))/((e^x-1)) ?

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

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inverse of ((e^x+1))/((e^x-1))

inverse

\frac{( e ^{x} + 1 )}{( e ^{x} - 1 )}

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

Solution steps

y = \frac{e ^{x} + 1}{e ^{x} - 1}

Replace

x

with

y

x = \frac{e ^{y} + 1}{e ^{y} - 1}

Solve for

y, x = \frac{e ^{y} + 1}{e ^{y} - 1}

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

f (x) = e^{\frac{x}{4.48}}

x^{0 - 3} + 1

\frac{x + 3}{x + 9}

\frac{+ z}{( \frac{1}{z} )}

f (x) = g (x) = (x + 2)^{3}

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