Single Variable Calculus, Chapter 7, 7.3-1, Section 7.3-1, Problem 34

Evaluate $\ln (2x + 1) = 2 - \ln x$ for $x$.

$
\begin{equation}
\begin{aligned}
\ln (2x + 1) &= 2 - \ln x\\
\\
\ln (2x + 1) + \ln x &= 2
\\
\ln [x (2x + 1)] &= 2\\
\\
e^{ \ln [x (2x + 1)] } &= e^2\\
\\
x(2x+ 1) &= e^2\\
\\
2x^2 + x &= e^2\\
\\
2x^2 + x - e^2 &= 0
\end{aligned}
\end{equation}
$


Using Quadratic Formula

$
\begin{equation}
\begin{aligned}
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
x &= \frac{-1 \pm \sqrt{(1)^2 - 4(2)(-e^2)}}{2(2)}\\
\\
x &= \frac{-1 \pm \sqrt{1+8e^2}}{4}
\end{aligned}
\end{equation}
$


Since, the value of $\displaystyle x = \frac{-1-\sqrt{1+8e^2}}{4}$ will result to a negative value it will not satisfy the original equation because $\ln - x$ is undefined. So $\displaystyle x = \frac{-1 + \sqrt{1+8e^2}}{4}$