From the given problem :
[tex]\ln (-x+1)-\ln (3x+5)=\ln (-6x+1)[/tex]Note that :
[tex]\ln M-\ln N=\ln \frac{M}{N}[/tex]The equation will be :
[tex]\begin{gathered} \ln \frac{-x+1}{3x+5}=\ln (-6x+1) \\ \frac{-x+1}{3x+5}=-6x+1 \\ -x+1=(-6x+1)(3x+5) \\ -x+1=-18x^2-27x+5 \\ 18x^2+26x-4=0 \\ 9x^2+13x-2=0 \end{gathered}[/tex]Using quadratic Formula :
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]a = 9, b = 13 and c = -2
[tex]\begin{gathered} x=\frac{-13\pm\sqrt[]{13^2-4(9)(-2)}}{2(9)} \\ x=\frac{-13\pm\sqrt[]{241}}{18} \\ x=0.14 \\ \text{and} \\ x=-1.58 \end{gathered}[/tex]The answer is :
x = 1.58 and 0.14