Question

If $2{\log }_{4}x=1+{\log }_4(x-1),$ find the value of $x$.

Answer 1

$2{\log }_{4}x=1+{\log }_4(x-1),$

$\Rightarrow {\log }_4{x}^2-{\log }_4(x-1)=1$

$\Rightarrow {\log }_4\frac{x^2}{x-1}=1$

$\Rightarrow 4^1=\frac{x^2}{x-1}$

$\Rightarrow x^2=4x-4$

$\Rightarrow x^2=4x+4=0$

$\Rightarrow (x-2{)}^2=0$

$\Rightarrow x=2$