# Find the solution of the equation : log_3{(x-3)+x^2-3x} = log _3(x-1)+log_3(x-3)?

Mar 15, 2018

$\phi$

#### Explanation:

$\textcolor{red}{\left(1\right) {\log}_{a} X = {\log}_{a} Y \implies X = Y}$
$\textcolor{red}{\left(2\right) {\log}_{a} M + {\log}_{a} N = {\log}_{a} \left(M N\right)}$
So
${\log}_{3} \left(x - 3 + {x}^{2} - 3 x\right) = {\log}_{3} \left(x - 1\right) + {\log}_{3} \left(x - 3\right) , . . u s e \left(2\right)$
${\log}_{3} \left({x}^{2} - 2 x - 3\right) = {\log}_{3} \left(\left(x - 1\right) \left(x - 3\right)\right)$
${x}^{2} - 2 x - 3 = \left(x - 1\right) \left(x - 3\right) , . . u s e \left(1\right)$
${x}^{2} - 2 x - 3 = {x}^{2} - 4 x + 3$
$- 2 x - 3 = - 4 x + 3$
$4 x - 2 x = 3 + 3 \implies 2 x = 6 \implies x = 3$
Check:
$L H S = {\log}_{3} \left(3 - 3 + 9 - 9\right) = {\log}_{3} \left(0\right) \to$undefined
$R H S = {\log}_{3} \left(3 - 1\right) + {\log}_{3} \left(3 - 3\right) = \log 2 + {\log}_{3} \left(0\right) \ne L H S$
$x = 3$ does not satisfy the equation.
OR
(x-3)+x^2-3x=(x-3)+x(x-3)=(x-3)(x+1)=>log_3(x-3)(x+1))=log_3(x-1)+log_3(x-3)=>log_3(x-3)+log_3(x+1)=log_3(x-1)+log_3(x-3)=>log_3(x+1)=log_3(x-1)
$\implies x + 1 = x - 1 \implies 1 = - 1$, which is not possible.