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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = x^{3} - 6 x^{2}

y = - \frac{4}{5} x + \frac{9}{5}, (- 5, - 5)

f (x) = x ln (x)

r (n) = \frac{1}{n - 1}

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