Mathematics (Factoring and Logarithms)

This is a discussion on Mathematics (Factoring and Logarithms) within the A Brief History of Cprogramming.com forums, part of the Community Boards category; I have a question. My book does not cover this, and I have been unable to locate anything on this ...

  1. #1
    essence of digital xddxogm3's Avatar
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    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
    "Hence to fight and conquer in all your battles is not supreme excellence;
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  2. #2
    & the hat of GPL slaying Thantos's Avatar
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    no you can not factor out like that.

  3. #3
    essence of digital xddxogm3's Avatar
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    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?
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

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    C++ Developer XSquared's Avatar
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    There is no solution to ln(2x)=ln(3x). ln(3x)-ln(2x) is approximately 0.4054651084 for all x.
    Last edited by XSquared; 11-25-2004 at 05:47 PM.
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  5. #5
    essence of digital xddxogm3's Avatar
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    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?
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  6. #6
    & the hat of GPL slaying Thantos's Avatar
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    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

  7. #7
    Rad gcn_zelda's Avatar
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    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.

  8. #8
    essence of digital xddxogm3's Avatar
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    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.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

  9. #9
    Rad gcn_zelda's Avatar
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    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.

  10. #10
    I'm less than sure.... abyssphobia's Avatar
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    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...
    [edit]

    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
    Last edited by abyssphobia; 11-25-2004 at 11:15 PM.
    Have I crossed the line?

  11. #11
    C++ Developer XSquared's Avatar
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    You're interpreting that property incorrectly. It would be ln(x/x)=ln(3/2), which gives 0=0.4054651084.
    Naturally I didn't feel inspired enough to read all the links for you, since I already slaved away for long hours under a blistering sun pressing the search button after typing four whole words! - Quzah

    You. Fetch me my copy of the Wall Street Journal. You two, fight to the death - Stewie

  12. #12
    Toaster Zach L.'s Avatar
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    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).
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  13. #13
    & the hat of GPL slaying Thantos's Avatar
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    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)

    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.

  14. #14
    Toaster Zach L.'s Avatar
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    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.
    The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.

  15. #15
    & the hat of GPL slaying Thantos's Avatar
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    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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