If then in order for a real solution to exist. In other words you discard -4 as a solution because it's not in the domain of the original equation
It works in because both x+2 and x-1 are negative when x=-4 then their product is positive.
Solving a Logarithmic Equation
log(x+2)+log(x-1)=1
the answers for this is x=-4 , x=3
x=-4 is undefined , however when we combine it
it becomes
log((x+2)(X-1))=1 and substitute x=-4 its defined .
Sense log(x+2)+log(x-1)=log((x+2)(X-1)) why there is a difference??
If then in order for a real solution to exist. In other words you discard -4 as a solution because it's not in the domain of the original equation
It works in because both x+2 and x-1 are negative when x=-4 then their product is positive.
In going from log(x+2)+log(x-1)=1 to log[(x+2)(x-1)]=1 you have changed the original equation and picked up an extraneous solution. A 'solution' found in this way must always be substituted into the original equation, checked and rejected (with reason given) if it does not satisfy the original equation.