Respuesta :
SOLUTION
We want to solve
[tex]\ln \mleft(-x+1\mright)-\ln \mleft(3x+5\mright)=\ln \mleft(-6x+1\mright)[/tex]This becomes
[tex]\begin{gathered} \ln (-x+1)-\ln (3x+5)=\ln (-6x+1) \\ \ln (-x+1)-\ln (3x+5)-\ln (-6x+1)=0 \\ \text{From laws of logarithm } \\ \ln \frac{(-x+1)}{(3x+5)}\times\frac{1}{(-6x+1)} \\ \ln \frac{(-x+1)}{(3x+5)(-6x+1)}=0 \end{gathered}[/tex]Since ln 1 = 0, we have
[tex]\begin{gathered} \ln \frac{(-x+1)}{(3x+5)(-6x+1)}=\ln 1 \\ \text{Canceling }\ln ,\text{ we have } \\ \frac{(-x+1)}{(3x+5)(-6x+1)}=1 \\ (-x+1)=(3x+5)(-6x+1) \end{gathered}[/tex]Expanding the equation in the right hand side we have
[tex]\begin{gathered} (-x+1)=(3x+5)(-6x+1) \\ (-x+1)=-18x^2+3x-30x+5 \\ -x+1=-18x^2+3x-30x+5 \\ -18x^2+3x-30x+5=-x+1 \\ -18x^2+3x-30x+x+5-1=0 \\ -18x^2-26x+4=0 \end{gathered}[/tex]Solving the quadratic equation we have
[tex]-18x^2-26x+4=0[/tex]We have
[tex]x=-1.58\text{ or }x=0.14[/tex]Hence, the answer is option 2
