# Mathematics (Factoring and Logarithms)

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• 11-25-2004
xddxogm3
Mathematics (Factoring and Logarithms)
I have a question. My book does not cover this, and I have been unable to locate anything on this online. Can you Factor inputs out of a logarithm?
Is the following valid? Please forgive the elementary question, but I feel it is important to get a good grasp on math to fine tune my code.
Code:

```ln(2x)=ln(3x) ln(2) + ln(x) = ln(3) + ln(x) ln(x) - ln(x) = ln(3) - ln(2) x ( ln() - ln() ) = ln (3/2) x = ln (3/2)```
if this is valid, what other properties/laws can I apply to logarithms?
associative, distributive, etc.
is there a list of these?
I found this site, but no answer to these questions.
http://mathworld.wolfram.com/Logarithm.html
• 11-25-2004
Thantos
no you can not factor out like that.
• 11-25-2004
xddxogm3
how would i issolate x in a problem like that?
w/o eliminating it totaly?
I'm assuming you can zero out the two ln(x) - ln(x).
and what properties can you apply to logarithms?
• 11-25-2004
XSquared
There is no solution to ln(2x)=ln(3x). ln(3x)-ln(2x) is approximately 0.4054651084 for all x.
• 11-25-2004
xddxogm3
I attended the tutorial session at my university, and the instructor in charge stated you can use the distributive law when dealing with logarithms. Is this not correct? If you can use distributive, wouldn't that allow the use of factoring out a common factor?
• 11-25-2004
Thantos
Code:

```ln(2x)=ln(3x) ln(2) + ln(x) = ln(3) + ln(x) ln(x) - ln(x) = ln(3) - ln(2)```
Up to this point you were ok.
Code:

`x ( ln() - ln() ) = ln (3/2)`
On the right side you were ok but on the left side you can not do that. This line should have looked like:
Code:

`0 = ln (3/2)`
which is a no solution as ln(3/2) does not equal 0
• 11-25-2004
gcn_zelda
ln(2x) = ln(3x)

Remove the natural log from both sides...

2x = 3x
x = 0

But you can't take a natural log of 0 because :

ln(0) is like saying e^? = 0
Nothing raised to a power can equal 0.
• 11-25-2004
xddxogm3
I found the name of the properties I was questioning.
Properties of Real Numbers.
I was advised by an instructor in a tutorial session, that you can use the Commutative, Associative, and Distributive properties on logarithms. Is this correct?
If you can use the Distributive property, couldn't you reverse that property?
This would be similar to factoring out a common multiple of the product.
Is this not a valid statement? Again, I just want to know if this is possible for future knowledge.
• 11-25-2004
gcn_zelda
ln(x) - ln(x) != x(ln - ln)

ln(x) is not natural log times x. It's the natural log of x. Logarithms are not constants. They're operations.
• 11-25-2004
abyssphobia
Quote:

Originally Posted by gcn_zelda
ln(x) - ln(x) != x(ln - ln)

ln(x) is not natural log times x. It's the natural log of x. Logarithms are not constants. They're operations.
ln(2x) = ln(3x)

Remove the natural log from both sides...

2x = 3x
x = 0

But you can't take a natural log of 0 because :

ln(0) is like saying e^? = 0
Nothing raised to a power can equal 0.

I agree with your theory !!!

Code:

```ln(2x)=ln(3x) ln(2) + ln(x) = ln(3) + ln(x) ln(x) - ln(x) = ln(3) - ln(2)```
but i am wondering , if my hypotesise could be possible
what if...
Code:

```ln x/ ln x = ln3/ln2    // because of the property of logarithms 1= ln3/ln2```
Is that possible or I'm dreaming up...

Quote:

Originally Posted by XSquared
You're interpreting that property incorrectly. It would be ln(x/x)=ln(3/2), which gives 0=0.4054651084.

right!!!
it's more logical,
well at least I had the idea ;)
thanks XSquared, you cleared mind :D
• 11-25-2004
XSquared
You're interpreting that property incorrectly. It would be ln(x/x)=ln(3/2), which gives 0=0.4054651084.
• 11-25-2004
Zach L.
Your problem is that x is not in the domain of ln. That is why you come up with a non-sense result. Surely, the one solution of 2x=3x is x=0, but 0 is not in the domain of ln, so that equation ln(2x)=ln(3x) has no solution. The properties of logarithms only apply when the quantities in question are in the domain.

To convince yourself that 0 is not in the domain, look at the definition of ln(x). Integral{t=1 to t=x} of dt/t. If you just sketch it out, you see that this integral is undefined (i.e. the 'area under the curve' between 0 and 1 inclusive diverges sharply).
• 11-26-2004
Thantos
You don't need integrals to see the domain of ln x. ln x is just the inverse of e^x. e^x has a domain of (-inf, +inf) and a range of (0, +inf). Since e^x is a one to one type of function we don't have to restrict it to take the inverse. So its inverse will have a domain of (0, +inf) and a range of (-inf, +inf)

Quote:

From IM
How would factoring with logarithms work?
If at all. Would I have to resolve the logarithm first before any attempt to factor?
You can not "factor" out of a log just as you can't "factor" out the base of an exponent. ie:
Code:

`x^2 + x^3 != x ( 1^2 + 1^3)`
I think you are forgetting the major rule: Once you get an answer put it into the orginal equation and check the results.
• 11-26-2004
Zach L.
Thats another way of looking at it. What I prefer about the integral solution is that it more directly relates to computing a value of ln(x) than does looking at it as the inverse of e^x.
• 11-26-2004
Thantos
But log is defined to be the inverse. Also consdiering the OP's previous math based questions I would consider the use of integrals to be more confusing then helpful.
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