What is the inverse of (e^y+e^{-y})/2 ?

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

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inverse of (e^y+e^{-y})/2

inverse

\frac{e ^{y} + e ^{- y}}{2}

ln (y + y^{2} - 1), ln (y - y^{2} - 1)

Solution steps

y = \frac{e ^{y} + e ^{- y}}{2}

Replace

y

with

y

y = \frac{e ^{y} + e ^{- y}}{2}

Solve for

y, y = \frac{e ^{y} + e ^{- y}}{2}

ln (y + y^{2} - 1), ln (y - y^{2} - 1)

f (x) = \frac{2 x + 1}{x ^{2} + 5 x + 4}

\frac{1}{x ^{2} - 4}

f (x) = \frac{sin ( x )}{x}

y = cos (x + 1)

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