Question

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

Answer 1

`phi`

Explanation:

`color(red)((1) log_aX=log_aY=>X=Y)`
`color(red)((2) log_aM+log_aN=log_a(MN))`
So
`log_3(x-3+x^2-3x)=log_3(x-1)+log_3(x-3),.. use (2)`
`log_3(x^2-2x-3)=log_3((x-1)(x-3))`
`x^2-2x-3=(x-1)(x-3),..use(1)`
`x^2-2x-3=x^2-4x+3`
`-2x-3=-4x+3`
`4x-2x=3+3=>2x=6=>x=3`
Check:
`LHS=log_3(3-3+9-9)=log_3(0) to`undefined
`RHS=log_3(3-1)+log_3(3-3)=log2+log_3(0)!=LHS`
`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)`
`=>x+1=x-1=>1=-1`, which is not possible.