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

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

inverse

(\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 )}

f (x) = 5 - lo g_{2} (3 x - 4)

f (x) = \frac{7 x - 8}{2 x + 9}

cos (9 0^{\circ})

f (x) = \frac{2 x + 3}{3 x - 4}

f (x) = \frac{5}{7} x + 4

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