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