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