To solve this question, we will proceed thus:
[tex]\ln (x)+\ln (x-1)=\ln (4x)[/tex]Simplifying further: (Applying log rules)
[tex]x(x-1)=4x_{}[/tex][tex]x^2-x=4x[/tex][tex]\begin{gathered} x^2-x-4x=0 \\ x^2-5x=0 \\ x(x-5)=0 \\ \text{This means that:} \\ x=0\text{ or } \\ x=5 \\ \\ To\text{ verify our solutions, we will put baxk the value of x into the initital } \\ expression\colon \\ \text{When x=0} \\ \ln (0)+\ln (0-1)=\ln (0)\text{ will result in math error:} \\ \text{When x=5} \\ \ln (5)+\ln (5-1)=\ln (4\times5)\text{ this will result in actual solutions} \\ \\ So\text{ with this above test, we can conclude that the correct solution is} \\ x=5 \\ \text{The correct answer therefore is C.} \end{gathered}[/tex]Answer = Option C.
