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

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

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inverse of f(x)=e^{(x+1)/(x-1)}

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{10 x + 5}{2 x - 10}

f (x) = g (x) = x^{2} - 5 x - 6

y = 10 - 2 x

f (x) = f (x) = 2 (x - 3) 2 + 5

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