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

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

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inverse of 2^{x^2-1}

inverse

2^{x^{2} - 1}

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

Solution steps

y = 2^{x^{2} - 1}

Replace

x

with

y

x = 2^{y^{2} - 1}

Solve for

y, x = 2^{y^{2} - 1}

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

11 cos (2 x) + 5

\frac{2 s + 3}{s ( 2 s + 1 )}

f (x) = \frac{1}{7} x^{2} + 3

f (x) = x^{2 - 10 + 15}

+ \frac{z ^{- 1}}{( 1 - 0.5 \cdot z ^{- 1} ) ^{2}}

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