From Förberedande kurs i matematik 1

If we write the equation as

\displaystyle \bigl(e^{x}\bigr)^{2} + e^{x} = 4

we see that \displaystyle x appears only in the combination \displaystyle e^{x} and it is therefore appropriate to treat \displaystyle e^{x} as a new unknown in the equation and then, when we have obtained the value of \displaystyle e^{x}, we can calculate the corresponding value of \displaystyle x by simply taking the logarithm.

For clarity, we set \displaystyle t=e^{x}, so that the equation can be written as

\displaystyle t^{2}+t=4

and we solve this second-degree equation by completing the square,

\displaystyle t^{2}+t = \Bigl( t+\frac{1}{2} \Bigr)^{2}-\Bigl( \frac{1}{2} \Bigr)^{2} = \Bigl( t+\frac{1}{2} \Bigr)^{2} - \frac{1}{4}\,,

which gives

\displaystyle \Bigl(t+\frac{1}{2}\Bigr)^{2} - \frac{1}{4} = 4\quad \Leftrightarrow \quad t = -\frac{1}{2}\pm \frac{\sqrt{17}}{2}\,\textrm{.}

These two roots give us two possible values for \displaystyle e^{x},

\displaystyle e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}\qquad\text{or}\qquad e^{x} = -\frac{1}{2}+\frac{\sqrt{17}}{2}\,\textrm{.}

In the first case, the right-hand side is negative and because "e raised to anything" can never be negative, there is no x that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because \displaystyle \sqrt{17}>1) and we can take the logarithm of both sides to obtain

\displaystyle x=\ln \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\,\textrm{.}

Note: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting \displaystyle t=\sqrt{17}/2-1/2 into the equation \displaystyle t^2+t=4,

\displaystyle \begin{align} \text{LHS} &= \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)^2 + \Bigl(\frac{\sqrt{17}}{2}-\frac{1}{2}\Bigr)\\[5pt] &= \frac{17}{4}-2\cdot \frac{1}{2}\cdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2}\\[5pt] &= \frac{17}{4}+\frac{1}{4}-\frac{1}{2}\\[5pt] &= \frac{17+1-2}{4}\\[5pt] &=\frac{16}{4}\\[5pt] &= 4\\[5pt] &= \text{RHS.} \end{align}