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

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

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

inverse

\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)^{\frac{1}{2}}

(- 7, - 7), (- 5, 5)

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

46

\frac{1}{x ^{3}}

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