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Funny logs

  1. Mar 30, 2007 #1
    Say you have the function f(x)=log(x).

    If x had dimensions of metres, say, then you can't have x=2m or whatever because dimensions can't be put into the equation to give log(2m).

    What about if you have x1=4m and x2=2m and put
    Each term doesn't exist by itself, but using log laws, this would be log(4m/2m)=log(2).
    So how can they exist together but not by themself?
  2. jcsd
  3. Mar 30, 2007 #2
    That's a good question!

    I don't have the answer- but note that


    So the units cancel when you write it this way too- as they should.
  4. Mar 30, 2007 #3
    okay how about 0 exist but 0/0 does not? or, 3+4i and 5+4i are not real but (5+4i) - (3+4i) = 2 is?
  5. Mar 30, 2007 #4
    that's true. Sill seems a bit strange though.
  6. Mar 30, 2007 #5
    log(4m/1m)-log(2m/1m) makes sense


    log(4m/1m)-log(2m/1m) = log(4m/2m)

    no mystery
  7. Mar 30, 2007 #6
    no? so what exactly log(2m) is?
  8. Mar 30, 2007 #7
    Correct me if I am incorrect.

    log(Cm) = log(C) + log(m) = log(C) + 0 = log(C)
  9. Mar 30, 2007 #8
    how do you know wheather log(m) is 0 or not??
  10. Mar 30, 2007 #9
    However this is not exactly what theperthvan asked!!
  11. Mar 30, 2007 #10
    so Cm = C and m = 1. and why we even need a word meters.
  12. Mar 30, 2007 #11
    I am using the logic that a meter is 1. By asking what log(m) is, I must ask what a meter is. In the mathematics a meter is basically just 1. By displaying 1 by an m we make it clear that this 1 is refering to a physical object.
  13. Mar 30, 2007 #12
    yeah, as m is merely a unit i think it should not need be written at all.
  14. Mar 30, 2007 #13
    Nope, the argument must be dimensionless. You can't just make a dimension a number.

    yeah, but then you get log(4m)-log(1m)-log(2m)+log(1m) which is back to the original question.

    Pretty much, the argument must be dimensionless, because the Taylor expansion will have x+x^2+x^3+... (well not for this function obviously, but the idea is the same). So that is equivalent to adding say 1m+1m^2+1m^3. i.e you cannot add length to area to volume to whatever comes next.
  15. Mar 30, 2007 #14
    maybe you would get dimension units in Taylor coefficients.
  16. Mar 30, 2007 #15


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    Understand that "meters" or "yards" or whatever are NOT "mathematical" entities. They are rather, parts of applications of mathematics to physics, geography, or whatever. The question you should be asking is not "what does log(3 meters) mean?" but rather "Are the any physics (or geography or...) formulas that require the logarithm of a distance?"- and if so, what does the physics say it means?

    I know several formula that use logarithms but they all take the logarithm of a ratio so that the units cancel.

    Let me ask you this: suppose on a quiz, you have a problem that defines f(x) to be x2 and then asks you to find f(3). You think "no problem, whip out your calculator and find that f(3)= 32= 9. On the very next question, g(x) is defined to be sin(x) and now you are asked to find g(3). You take your calculator, enter sin(3) and it gives you 0.0532359562 (approximately).

    But when you get the quiz paper back, you find that problem has been marked wrong and your instructor tells you that the correct answer is 0.1411200081 and that you should have had your calculator in "radian mode" instead of "degree mode"! Should you complain? Go to the department chair? The president? The police? The problem didn't say anything about "degrees" or "radians"! Shouldn't either answer be acceptable?

    Of course, the first question, "find f(3) when f(x)= x2" didn't say anything about units either. Mathematical functions deal with numbers, not measurements and so not units. That's one reason why the sine and cosine functions used in Calculus and above are not defined in terms of right triangles.
  17. Mar 30, 2007 #16
    Though, implementing my logic again, m with dimensions is just 1 in that dimension. m^2 is just 1 squared area. m^3 is just 1 cube.
  18. Mar 30, 2007 #17
    Also, log_a(x) is the exponent of base a. Thus, when we say log_a(4m) and a = 2 (for ease of use), implementing that logic that m is 1 to obtain the exponent, it functions correctly when producing x.

    log_2(4m) = log(4) = 2

    2^2m = 4m

    Thus, it functions fine.
  19. Mar 30, 2007 #18

    Gib Z

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    Homework Helper

    The only reason it functions fine is because you are treating m as a number! For all we know it is an unknown value, so the logarithm operator can be applied to it. However you are using it as meters, a dimension, but operating on it like a number!
  20. Apr 1, 2007 #19
    what a f????

    by definition of units, we have equations like 1 * meter = 39.3700787 * inch, or 1 hectare = 10 000 meter * meter. please show me how do you arrive at 2^(2 * meter) = 4 * meter from there?
  21. Apr 1, 2007 #20

    Gib Z

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    Ahh no you misinterpreted that quote, not at your own fault however. The notation was ambiguous. He meant (2^2)m=4m, not 2^(2m)=4m.
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