Answer
$x = \frac{5+\sqrt {37}}{2}$
Work Step by Step
$\ln (2x+1) + \ln (x-3) - 2\ln x = 0$
$\ln (2x+1)(x-3) = 2\ln x$
$\ln (2x+1)(x-3) = \ln x^{2}$
$(2x + 1)(x-3) = x^{2}$
$2x(x-3)+1(x-3) = x^{2}$
$2x^{2} - 6x + x - 3 = x^{2}$
$x^{2} - 5x - 3 = 0$
$x = \frac{-b±\sqrt {b^{2}-4ac}}{2a}$
$x = \frac{-(-5)±\sqrt {(-5)^{2}-4(1)(-3)}}{2(1)}$
$x = \frac{5±\sqrt {25+12}}{2}$
$x = \frac{5±\sqrt {37}}{2}$
Since $x \ne \frac{5-\sqrt {37}}{2}$, then $x = \frac{5+\sqrt {37}}{2}$
