## Precalculus (6th Edition) Blitzer

The solution set is $\displaystyle \{\frac{5+\sqrt{37}}{2}\}.$
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 $\displaystyle \{\frac{5+\sqrt{37}}{2}\}.$