What is the inverse of f(x)=ln(e^x-1)-ln(2)-1 ?

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

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inverse of f(x)=ln(e^x-1)-ln(2)-1

inverse

f (x) = ln (e^{x} - 1) - ln (2) - 1

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{1}{x + 14}

f (x) = x^{2} - 4 x + 7

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

cos (x) + sin (x)

(x + 3) (- x^{2} + 16)

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