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

1 Answer
maganbhai P.
Mar 15, 2018

#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.

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