What is the inverse of y=ln(x-sqrt(x^2-1)) ?

The inverse of y=ln(x-sqrt(x^2-1)) is -(-1-e^{2x})/(2e^x)

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inverse of y=ln(x-sqrt(x^2-1))

inverse

y = ln (x - x^{2} - 1)

- \frac{- 1 - e ^{2 x}}{2 e ^{x}}

Solution steps

y = ln (x - x^{2} - 1)

Replace

x

with

y

x = ln (y - y^{2} - 1)

Solve for

y, x = ln (y - y^{2} - 1)

- \frac{- 1 - e ^{2 x}}{2 e ^{x}}

\frac{3 - 3 x}{x - 5}

6 x - 6

f (x) = (e^{x^{2}})^{+ 1}

\frac{2 x ^{2} + 1}{3 x - 5}

\frac{5 x + 5}{8 x + 32}

What is the inverse of y=ln(x-sqrt(x^2-1)) ?
The inverse of y=ln(x-sqrt(x^2-1)) is -(-1-e^{2x})/(2e^x)