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)}

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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}}

\frac{x - 1}{3 - 2 x}

f (x) = 3 x^{2} - 7

f (x) = 1 - \frac{1}{x ^{2}}

f (x) = \frac{5 - 4 x}{3 x}

f (x) = lo g_{2} (5 x - 2)

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)}