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Logarithm problem

by Bassalisk
Tags: logarithm
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Bassalisk
#1
Jul5-11, 01:10 PM
P: 951
I have a strange problem.

Have this example.

log(x-2)+log(x+1)=0

Domain of these functions are: x>2, x>-1 resulting in x>2;

by logarithm rule we can combine these 2 logarithms and make one logarithm.

log(x-2)(x+1)=0

(-infinity,-1) U (2,+infinity)

But the domain of this changes. How can we combine these 2 logarithms if the domain changes? Am I missing something here?
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I like Serena
#2
Jul5-11, 01:44 PM
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In your original problem you have 2 log functions.
The domain of both needs to be respected for any solution to be a proper solution.

When you multiply the 2 factors, you effectively extend the domain.
This is because both factors might be negative, yielding a positive number that is inside the domain of the log function.

However, solutions for which these numbers are negative do not satisfy the original problem and will have to be discarded.
Bassalisk
#3
Jul5-11, 02:06 PM
P: 951
Quote Quote by I like Serena View Post
In your original problem you have 2 log functions.
The domain of both needs to be respected for any solution to be a proper solution.

When you multiply the 2 factors, you effectively extend the domain.
This is because both factors might be negative, yielding a positive number that is inside the domain of the log function.

However, solutions for which these numbers are negative do not satisfy the original problem and will have to be discarded.
Yes but how can logarithm rule then stand if the domain changes, shouldn't I change the domains too?

log(x-2)+log(x+1)=log(x-2)(x+1) how can this equality hold?

I like Serena
#4
Jul5-11, 02:11 PM
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Logarithm problem

The equality only holds for values of x with which the domains of all log functions in the expression are satisfied.
Otherwise it is undefined.
Bassalisk
#5
Jul5-11, 02:12 PM
P: 951
Quote Quote by I like Serena View Post
The equality only holds for values of x with which the domains of all log functions in the expression are satisfied.
Otherwise it is undefined.
So basically yes, I do have to change the domain.
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#6
Jul5-11, 02:36 PM
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Quote Quote by Bassalisk View Post
So basically yes, I do have to change the domain.
Uhh, noooo!
At the start of the problem, the domain is (2, +infinity).
This never changes throughout the problem.
Bassalisk
#7
Jul5-11, 02:40 PM
P: 951
Quote Quote by I like Serena View Post
Uhh, noooo!
At the start of the problem, the domain is (2, +infinity).
This never changes throughout the problem.
Thank you, I understand now ^^
BobG
#8
Jul5-11, 03:06 PM
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Quote Quote by Bassalisk View Post
I have a strange problem.

Have this example.

log(x-2)+log(x+1)=0

Domain of these functions are: x>2, x>-1 resulting in x>2;

by logarithm rule we can combine these 2 logarithms and make one logarithm.

log(x-2)(x+1)=0

(-infinity,-1) U (2,+infinity)

But the domain of this changes. How can we combine these 2 logarithms if the domain changes? Am I missing something here?
ILike Serena's explanation is correct. The domain won't change, since you're dealing with real numbers.

However, you illustrate and interesting point. If x = -2, then the complex solution is:

ln (-4) + ln (-1) = ln (-4*-1)
(1.386,pi) + (0,pi) = (1.386,2pi)

And, if you start at 0 and go 2pi radians around a circle, you wind up back at 0. The discrepancy is because you're restricting yourself solely to the real numbers. (Plus, the result is only meaningful when you're using natural logarithms.)

So, suffice it to say, x has to be greater than 2 and that won't change when you combine them.
Bassalisk
#9
Jul6-11, 03:59 AM
P: 951
But what if I start with the problem log(x-2)(x+1)=0 and I divide into 2 parts,

log(x-2)+log(x+1)=0. Do I have to change the domain then?

because domain of the first is
(-infinity,-1) U (2,+infinity) and the domain after the transformation is (2,+infinity)
I like Serena
#10
Jul6-11, 04:21 AM
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This would mean that you have to distinguish two cases: the case x < -1 and the case x > 2.

In the case x < -1 you would get:
log(-(x-2)) + log(-(x+1)) = 0
which you have to solve separately.

When you have solved both cases, you need to combine all the solutions that you found.
Bassalisk
#11
Jul6-11, 04:23 AM
P: 951
Thank you, I totally forgot about that. Mind filled with transistors and diodes :p


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