Solve for x y=e^(-x^2)

$y = e^{- x^{2}}$

Rewrite the equation as $e^{- x^{2}} = y$ .

$e^{- x^{2}} = y$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- x^{2}}) = \ln (y)$

Expand the left side.

Expand $\ln (e^{- x^{2}})$ by moving $- x^{2}$ outside the logarithm.

$- x^{2} \ln (e) = \ln (y)$

The natural logarithm of $e$ is $1$ .

$- x^{2} \cdot 1 = \ln (y)$

Multiply $- 1$ by $1$ .

$- x^{2} = \ln (y)$

Divide each term in $- x^{2} = \ln (y)$ by $- 1$ and simplify.

Divide each term in $- x^{2} = \ln (y)$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Move the negative one from the denominator of $\frac{\ln (y)}{- 1}$ .

$x^{2} = - 1 \cdot \ln (y)$

Rewrite $- 1 \cdot \ln (y)$ as $- \ln (y)$ .

$x^{2} = - \ln (y)$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- \ln (y)}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{- \ln (y)}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{- \ln (y)}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{- \ln (y)}$

$x = - \sqrt{- \ln (y)}$

$x = \sqrt{- \ln (y)}$

$x = - \sqrt{- \ln (y)}$