From Förberedande kurs i matematik 1

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

\displaystyle \begin{align} \text{LHS} &= \ln 2^{-x^{2}} = -x^{2}\cdot \ln 2\,,\\[5pt] \text{RHS} &= \ln \bigl(2e^{2x}\bigr) = \ln 2 + \ln e^{2x} = \ln 2 + 2x\cdot \ln e = \ln 2 + 2x\cdot 1\,\textrm{.} \end{align}

After a little rearranging, the equation becomes

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

We complete the square of the left-hand side,

\displaystyle \Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} - \Bigl(\frac{1}{\ln 2} \Bigr)^{2} + 1 = 0\,,

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

\displaystyle \Bigl(x+\frac{1}{\ln 2}\Bigr)^{2} = \Bigl(\frac{1}{\ln 2} \Bigr)^{2} - 1\,\textrm{.}

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 \ln 2 < \ln e = 1\,, we must have that \displaystyle (1/\ln 2)^{2} > 1\,, i.e. the right-hand side is positive.

The equation therefore has the solutions

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

which can also be written as

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