Answer
The solution set is $\displaystyle \{\frac{5+\sqrt{37}}{2}\}.$
Work Step by Step
First, in order for the equation to be defined, $\left[\begin{array}{lll}
2x+1 \gt 0 & \Rightarrow & x \gt -1/2\\
x-3 \gt 0 & \Rightarrow & x \gt 3\\
x\gt 0 & &
\end{array}\right] \Rightarrow x \gt 3 \quad (*)$
$\ln(2x+1)+\ln(x-3)-2\ln x=0$
Apply the product and power rules
$\ln[(2x+1)(x-3)]-\ln x^{2}=0$
Apply the quotient rule, ... $0=\ln 1$
$\displaystyle \ln\frac{(2x+1)(x-3)}{x^{2}}=\ln 1$
Logarithmic functions are one-to-one
$\displaystyle \frac{(2x+1)(x-3)}{x^{2}}=1$
$2x^{2}-5x-3=x^{2}$
$ x^{2}-5x-3=0$
Use the quadratic formula $ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$ x=\displaystyle \frac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(-3)}}{2(1)}=\frac{5\pm\sqrt{37}}{2}$
$ x=\displaystyle \frac{5+\sqrt{37}}{2}\approx 5.54\quad $ satisfies (*)
$ x=\displaystyle \frac{5-\sqrt{37}}{2}\approx-0.54 \quad $ does not satisfy (*), discard.
The solution set is $\displaystyle \{\frac{5+\sqrt{37}}{2}\}.$
