Question

Solve the following equation.
${\log }_4(x^2-1)-{\log }_4(x-1{)}^2={\log }_4\sqrt{(4-x{)}^2}$

Answer 1

${\log }_4(x^2-1)-{\log }_4({x-1)}^2={\log }_4\sqrt{{(4-x)}^2}\Rightarrow {\log }_4\frac{(x^2-1)}{{(x-1)}^2}={\log }_4\sqrt{{(4-x)}^2}\Rightarrow \frac{x^2-1}{{(x-1)}^2}=4-xorx-4.Firststep\frac{x+1}{x-1}=4-x,x\ne 1\Rightarrow x+1={-x}^2+5x-4\Rightarrow -x^2+4x-5=0\Rightarrow x^2-4x+5=0Nosolution.SecondStep.\frac{x+1}{x-1}=x-4\Rightarrow x+1=x^2-5x+4\Rightarrow x^2-6x+3=0\Rightarrow x=3\pm \sqrt{6}.Butx\ne 3-\sqrt{6}.Therefore,x^2-1<0atx=3-\sqrt{6}Therefore,x=3+\sqrt{6}isonlysolution.$