Aus Online Mathematik Brückenkurs 1

If we write the equation as

\displaystyle \left( e^{x} \right)^{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=\left( t+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}

which gives

\displaystyle \left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}=4\quad \Leftrightarrow \quad t=-\frac{1}{2}\pm \frac{\sqrt{17}}{2}

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

\displaystyle e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2} or \displaystyle e^{x}=-\frac{1}{2}+\frac{\sqrt{17}}{2}

In the first case, the right-hand side is negative and because " \displaystyle e raised to anything" can never be negative, there is no \displaystyle 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 \left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)

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

LHS \displaystyle =

\displaystyle \begin{align} & =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\ & =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\ \end{align}

\displaystyle = RHS.