What is the inverse of f(x)=log_{2}((e^x-1)/(-2)) ?

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

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inverse of f(x)=log_{2}((e^x-1)/(-2))

inverse

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

ln (- 2^{x + 1} + 1)

Solution steps

y = lo g_{2} (\frac{e ^{x} - 1}{- 2})

Replace

x

with

y

x = lo g_{2} (\frac{e ^{y} - 1}{- 2})

Solve for

y, x = lo g_{2} (\frac{e ^{y} - 1}{- 2})

ln (- 2^{x + 1} + 1)

5 x^{3} - 8

g (x) = x + 1

f (x) = \frac{1}{2} x^{3} + 1

f (x) = \frac{1}{2} x^{3} + 3

\frac{2 x ^{2} + 14 x - 35}{x + 9}

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