Solve for x 13e^(x^2+3)=39

$13 e^{x^{2} + 3} = 39$

Divide each term in $13 e^{x^{2} + 3} = 39$ by $13$ and simplify.

Divide each term in $13 e^{x^{2} + 3} = 39$ by $13$ .

$\frac{13 e^{x^{2} + 3}}{13} = \frac{39}{13}$

Simplify the left side.

Cancel the common factor of $13$ .

Cancel the common factor.

$\frac{13 e^{x^{2} + 3}}{13} = \frac{39}{13}$

Divide $e^{x^{2} + 3}$ by $1$ .

$e^{x^{2} + 3} = \frac{39}{13}$

Simplify the right side.

Divide $39$ by $13$ .

$e^{x^{2} + 3} = 3$

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

$\ln (e^{x^{2} + 3}) = \ln (3)$

Expand the left side.

Expand $\ln (e^{x^{2} + 3})$ by moving $x^{2} + 3$ outside the logarithm.

$(x^{2} + 3) \ln (e) = \ln (3)$

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

$(x^{2} + 3) \cdot 1 = \ln (3)$

Multiply $x^{2} + 3$ by $1$ .

$x^{2} + 3 = \ln (3)$

Subtract $3$ from both sides of the equation.

$x^{2} = \ln (3) - 3$

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

$x = \pm \sqrt{\ln (3) - 3}$

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.

$x = \sqrt{\ln (3) - 3}$

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

$x = - \sqrt{\ln (3) - 3}$

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

$x = \sqrt{\ln (3) - 3}, - \sqrt{\ln (3) - 3}$