Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 12

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



$
\begin{equation}
\begin{aligned}

& \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)}}
\\
\\
x =& \frac{-1 \pm \sqrt{1 + 8e^2}}{4}


\end{aligned}
\end{equation}
$


Since, $\displaystyle x = \frac{-1 - \sqrt{1 + 8e^2}}{4}$ will not satisfy the original equation because $\displaystyle \ln \frac{-1 - \sqrt{1 + 8e^2}}{4}$ is undefined.

Therefore, $\displaystyle x = \frac{-1 + \sqrt{1 + 8e^2}}{4}$