Solving Logarithm Problem Where Domain Changes

  • Thread starter Bassalisk
  • Start date
  • Tags
    Logarithm
In summary, when you divide a real number by a complex number, the domain of the real number doesn't change, but the domain of the complex number changes. This means that you have to solve two separate cases, depending on whether x is less than or greater than -1.
  • #1
Bassalisk
947
2
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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
I like Serena said:
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?
 
  • #4
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.
 
  • #5
I like Serena said:
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.
 
  • #6
Bassalisk said:
So basically yes, I do have to change the domain.

Uhh, noooo! :uhh:
At the start of the problem, the domain is (2, +infinity).
This never changes throughout the problem.
 
  • #7
I like Serena said:
Uhh, noooo! :uhh:
At the start of the problem, the domain is (2, +infinity).
This never changes throughout the problem.

Thank you, I understand now ^^
 
  • #8
Bassalisk said:
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.
 
  • #9
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)
 
  • #10
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.
 
  • #11
Thank you, I totally forgot about that. Mind filled with transistors and diodes :p
 

1. What is a logarithm?

A logarithm is the opposite operation of exponentiation. It is used to solve problems where the unknown is in the exponent.

2. What is the domain of a logarithm?

The domain of a logarithm is the set of values for which the logarithm function is defined. In other words, it is the set of all positive real numbers.

3. How does the domain change in logarithm problems?

In logarithm problems, the domain changes when the value inside the logarithm function becomes negative. This results in an undefined output.

4. How do you solve a logarithm problem with a changing domain?

To solve a logarithm problem with a changing domain, you can use the properties of logarithms to rewrite the problem into a form where the domain is valid. This may involve factoring, simplifying, or using the change of base formula.

5. Can a logarithm problem have multiple solutions?

Yes, a logarithm problem can have multiple solutions. This is because logarithms are not one-to-one functions, meaning that different inputs can result in the same output. It is important to check your solution by plugging it back into the original problem to ensure it is valid.

Similar threads

  • Calculus
Replies
4
Views
2K
Replies
3
Views
1K
  • Calculus
Replies
5
Views
11K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
10
Views
782
  • Precalculus Mathematics Homework Help
Replies
8
Views
600
Replies
5
Views
1K
Replies
1
Views
793
Back
Top