Question

Solve the following equation.
$2{\log }_3\frac{x-3}{x-7}+1={\log }_3\frac{x-3}{x-1}$

Answer 1

${2\log }_3\left(\frac{x-3}{x-7}\right)+1={\log }_3\left(\frac{x-3}{x-1}\right)\Rightarrow {\log }_3({\frac{x-3}{x-7})}^2+{\log }_{3}3={\log }_3(\frac{x-3}{x-1})\Rightarrow \frac{{3(x-3)}^2}{{(x-7)}^2}=\frac{x-3}{x-1}\Rightarrow \frac{3x-9}{{(x-7)}^2}=\frac{1}{x-1},x\ne 3{3x}^2-12x+9=x^2-14x+49\Rightarrow {2x}^2+2x-40=0\Rightarrow x^2+x-20=0\Rightarrow x=-5,4.Atx=4,\frac{x-3}{x-7}=\frac{1}{-3}<0Therefore,x=-5isonlysolution.$