# Thread: Logarithmic Equation is undefined (help).

1. ## Logarithmic Equation is undefined (help).

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??

2. ## Re: Logarithmic Equation is undefined (help).

If $y = \log[f(x)]$ then $f(x) > 0$ 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 $\log[(x+2)(x-1)]$ because both x+2 and x-1 are negative when x=-4 then their product is positive.

3. ## Re: Logarithmic Equation is undefined (help).

Originally Posted by e^(i*pi)
If $y = \log[f(x)]$ then $f(x) > 0$ 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 $\log[(x+2)(x-1)]$ because both x+2 and x-1 are negative when x=-4 then their product is positive.
But why do both give different answers while they are they same??

according to log(A.B)=log(A)+log(B)

log(x+2)+log(x-1)=log((x+2)(X-1))

4. ## Re: Logarithmic Equation is undefined (help).

Originally Posted by omar1
But why do both give different answers while they are they same??

according to log(A.B)=log(A)+log(B)

log(x+2)+log(x-1)=log((x+2)(X-1))
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.

5. ## Re: Logarithmic Equation is undefined (help).

Thank you very much for helping me!
Mr. fantastic and e^(i*pi) .