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

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

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

inverse

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

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

Solution steps

y = 2^{x + 1}

Replace

x

with

y

x = 2^{y + 1}

Solve for

y, x = 2^{y + 1}

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

\frac{x}{5} + 3

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

f (x) = 3 x^{2} + 30 x + 12

x^{2} + 12 x

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

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