From Förberedande kurs i matematik 1

Both left- and right-hand sides are positive for all values of \displaystyle x and this means that we can take the logarithm of both sides and get a more manageable equation.

LHS \displaystyle =\ln 2^{-x^{2}}=-x^{2}\centerdot \ln 2,

RHS \displaystyle =\ln \left( 2e^{2x} \right)=\ln 2+\ln e^{2x}=\ln 2+2x\centerdot \ln e=\ln 2+2x\centerdot 1

After a little rearranging, the equation becomes

\displaystyle x^{2}+\frac{2}{\ln 2}x+1=0

We complete the square of the left-hand side

\displaystyle \left( x+\frac{1}{\ln 2} \right)^{2}-\left( \frac{1}{\ln 2} \right)^{2}+1=0

and move the constant terms over to the right-hand side:

\displaystyle \left( x+\frac{1}{\ln 2} \right)^{2}=\left( \frac{1}{\ln 2} \right)^{2}-1

It can be difficult to see whether the right-hand side is positive or not, but if we remember that \displaystyle e>2 and that thus \displaystyle \text{ln 2}<\ln e=\text{1}, we must have that \displaystyle \left( {1}/{\ln 2}\; \right)^{2}>1, i.e. the right-hand side is positive.

The equation therefore has the solutions

\displaystyle x=-\frac{1}{\ln 2}\pm \sqrt{\left( \frac{1}{\ln 2} \right)^{2}-1,}

which can also be written as

\displaystyle x=\frac{-1\pm \sqrt{1-\left( \ln 2 \right)^{2}}}{\ln 2}