What is the inverse of 1/2 ln((1+x)/(1-x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of 1/2 ln((1+x)/(1-x))

inverse

\frac{1}{2} ln (\frac{1 + x}{1 - x})

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

Solution steps

y = \frac{1}{2} ln (\frac{1 + x}{1 - x})

Replace

x

with

y

x = \frac{1}{2} ln (\frac{1 + y}{1 - y})

Solve for

y, x = \frac{1}{2} ln (\frac{1 + y}{1 - y})

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

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