SOLUTION: Question1: log(x-2)+log(x+1)=2 This is what I have so far.. However, I'm not confident it's correct. Please help. Thank you!! log[(x-2)(x+1)]=2 b=1 p=2 x=(x-2)(x+1)

Question 378300: Question1: log(x-2)+log(x+1)=2
This is what I have so far.. However, I'm not confident it's correct. Please help. Thank you!!
log[(x-2)(x+1)]=2
b=1 p=2 x=(x-2)(x+1)

Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
log(x-2) + log(x+1) = 2
You are correct with
log((x-2)(x+1)) = 2
But I am confused with the "b=1 p=2 and x=(x-2)(x+1)"

The next step is to rewrite the equation in exponential form. In general is equivalent to . Using this on our equation we get:
(Since the base of log is 10.)
Simplifying each side we get:

This is a quadratic equation so we want one side to be zero. Subtracting 100 from each side we get:

This will not factor but we can use the Quadratic Formula:

which simplifies as follows:

In long form this is:
or

When solving equations where the variable is in the argument (or base) of a logarithm, you must check you answers. You must make sure that all arguments (and bases) of logarithms are positive. If a "solution" makes an argument (or base) negative or zero then that "solution" must be rejected. This can happen even if no mistakes were made while finding the solution! This is why we must check out answers in problems like this.

When checking answers use the original equation:
log(x-2) + log(x+1) = 2
Checking :

Since the we should be able to see that both arguments will be positive. So there is no reason to reject this solution. The rest of the check is optional. It will determine if we made a mistake. I'll leave that up to you.

Checking :

Since the we should be able to see that both arguments will be negative! So we must reject this "solution".

So the only solution to your equation is:

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