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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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