Solution 3.4:2b
From Förberedande kurs i matematik 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.
