# Logarithmic Equation is undefined (help).

• Aug 26th 2011, 01:13 PM
omar1
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??
• Aug 26th 2011, 01:30 PM
e^(i*pi)
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.
• Aug 26th 2011, 01:59 PM
omar1
Re: Logarithmic Equation is undefined (help).
Quote:

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))
• Aug 26th 2011, 02:21 PM
mr fantastic
Re: Logarithmic Equation is undefined (help).
Quote:

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.
• Aug 26th 2011, 06:29 PM
omar1
Re: Logarithmic Equation is undefined (help).
Thank you very much for helping me!
Mr. fantastic and e^(i*pi) .(Cool)(Bow)