On the left side, you can condense the logarithms into one:
[tex]\ln(1-x)-\ln(3x+5)=\ln\dfrac{1-x}{3x+5}[/tex]
Then
[tex]\ln\dfrac{1-x}{3x+5}=\ln(1-6x)\implies e^{\ln((1-x)/(3+5))}=e^{\ln(1-6x)}\implies\dfrac{1-x}{3x+5}=1-6x[/tex]
From here it's a purely algebraic equation. Multiply both sides by [tex]3x+5[/tex] to get
[tex]1-x=(1-6x)(3x+5)[/tex]
[tex]1-x=5-27x-18x^2[/tex]
[tex]18x^2+26x-4=0[/tex]
[tex]9x^2+13x-2=0[/tex]
By the quadratic formula,
[tex]x=\dfrac{-13\pm\sqrt{241}}{18}[/tex]
or about [tex]x\approx-1.5847[/tex] and [tex]x\approx0.14023[/tex].
Before we finish, first note that in order for the original equation to make sense, we need [tex]x[/tex] to satisfy 3 conditions:
[tex]-x+1>0\implies x<1[/tex]
[tex]3x+5>0\implies x>-\dfrac53\approx-1.67[/tex]
[tex]-6x+1>0\implies x<\dfrac16\approx0.17[/tex]
or taken together,
[tex]-\dfrac53<x<\dfrac16[/tex]
so both solutions found above are valid.
