## College Algebra (6th Edition)

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