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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

y = \frac{7 x + 5}{x - 1}

f (x) = 7 + ln (3 x + 1)

\frac{x + 1}{x + 9}

f (x) = 5000 e^{- 0.0005 x}

f (x) = ln (2 - x) - 1

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