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Proof of the logarithm product rule

  1. Jul 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that $$log_{b}(xy)=log_{b}x+log_{b}y.$$

    2. Relevant equations
    Let $$b^{u}=x,b^{v}=y.$$ Then $$log_{b}x=u,log_{b}y=v.$$

    3. The attempt at a solution
    I'm afraid I've been using circular reasoning to prove this. I can get this to a point where I have $$log_{b}(b^{u+v})=log_{b}b^{u}+log_{b}b^{v},$$ but I don't have a good way to simplify either side. There is a property which I later prove for which $$log_{b}(b)^{a}=a.$$ But I need the proof of the product rule in order to prove this! Every proof I've referred to uses the exponent rule (just shown) to prove the product rule, but... That doesn't seem quite right. What am I missing?
    Thanks for your time and help.
     
  2. jcsd
  3. Jul 21, 2015 #2

    Mark44

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    The left side simplifies to u + v = ##log_b(x) + log_b(y)##.
     
  4. Jul 21, 2015 #3
    How does the left side simplify to that? If I'm not mistaken, you've used the exponent property which states $$log_{b}b^{a}=a*log_{b}b=a*1=a,$$ which, as I've said, I cannot use because I've not proven that yet. Furthermore, in my proof of that exponent property, I use the product of logarithms property... But I can't do that, since that would be circular reasoning!
    Thanks again for your time.
    Joe
     
  5. Jul 21, 2015 #4

    SammyS

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    No, I don't think that's what Mark used. I think he used the following from you Original Post:
     
  6. Jul 21, 2015 #5
    He did say "left side," and my left side was $$log_{b}(b^{u+v}).$$ The right side $$log_{b}x+log_{b}y$$ can be simplified into $$u+v$$ by the fact that $$log_{b}x=u,log_{b}y=v,$$ (which I embarrassingly overlooked) but then that would leave me with $$log_{b}(b^{u+v})=u+v.$$

    Thanks again,
    Joe
     
  7. Jul 22, 2015 #6
    The "Relevant equations" section has enough information to simplify the left side without any circularity, and as SammyS has pointed out, that is all that was used. May I ask how you use the product rule to prove the exponent rule for logarithms?
     
  8. Jul 22, 2015 #7

    Fredrik

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    This is a bit of a nitpick, but it would be an improvement if the statement of the theorem mentions what values of x, y and b are allowed. The statement should be something like this:

    For all positive real numbers x,y,b we have ##\log_b(xy)=\log_bx+\log_by##.
    Then when you prove the theorem, you should start by saying that x,y and b satisfy the assumptions of the theorem, but are otherwise arbitrary. This is the simplest way to do that:

    Let x,y,b>0.
    You need to define the variables u and v, and I think the clearest way to do that is through the equations ##u=\log_b x## and ##v=\log_b y##. Note that these definitions wouldn't make sense if we didn't have x,y,b>0. That's why the sentence I suggested you put at the start of the proof is important. Now that you have defined u and v, you can say that the definitions imply that ##b^u= b^{\log_b x}=x## and ##b^v=b^{\log_b y}=y##. Now you're ready to do the the calcuations.

    Are you asking why this holds? It follows immediately from the definition of ##\log_b##. For each ##b>0##, ##\log_b## is defined as the inverse of the function ##f:\mathbb R\to(0,\infty)## defined by ##f(x)=b^x## for all real numbers x.
     
    Last edited: Jul 22, 2015
  9. Jul 22, 2015 #8
    Consider the function ##f(x) = \ln(xy) ## for ## x,y > 0 ##.
    Can you see that ## f'(x) = \ln' (x) ## ?
    That means that ##f(x)## and ##\ln(x)## vary by an additive constant right ? Show that this constant is ##\ln(y)##. Then it is easy to show the formula I think.
     
  10. Jul 22, 2015 #9
    Not a nitpick at all, it's something I do need to have. I just gave a briefer version here to save on the amount of time it takes me to post. Excellent advice though, and I agree with you completely.

    I see why $$log_{b}b$$ cancels, but it's not clear to me (unless using the exponent rule) that that exponent "comes down" to become a normal number. For example, when I see $$log_{b}(b^{u+v})$$ the definition of logarithms alone would get me $$1^{u+v}.$$ Obviously this isn't the case because of the exponent rule, but if I've not proven or even demonstrated how that works, what would lead me to believe that that is what will happen? I've shown how inverses, through composition, leave me with the variable x, i.e. $$f^{-1}(f(x))=x,$$ but again, this doesn't tell me what to do with the exponent.

    Absolutely. I'm not sure how to use Latex for this one, so I will share a .pdf of the proof on my OneDrive. Here is the link.

    Thanks to everyone for your help. I greatly appreciate you taking the time to help me out on this one.
    Joe
     
  11. Jul 22, 2015 #10

    Fredrik

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    You made a mistake here. The result is u+v, not ##1^{u+v}##. (See my next comment below).

    When f is defined as in my previous post, we have
    $$\log_b(b^{u+v})=\log_b(f(u+v))=f^{-1}(f(u+v))=u+v.$$
     
  12. Jul 22, 2015 #11

    Fredrik

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    This proof is fine when u is an integer, but you should be able to prove it when u is an arbitrary real number. Hint: Define ##v=\log_b a## so that ##b^v=a##. You will need to use the rule ##(a^x)^y=a^{xy}##.
     
  13. Jul 22, 2015 #12

    Mark44

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    This thread was posted in the Precalculus section, so it is likely that the OP has no idea what you're saying here.
     
  14. Jul 22, 2015 #13
    I hope he tells us because I was glad to have something to say
     
  15. Jul 22, 2015 #14

    Mark44

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    For a thread posted in the Precalc section, a reasonable (and usually correct) assumption is that the OP has not learned calculus yet. If the help you provide is way over the head of the person you're attempting to help, you haven't really helped that person.
     
  16. Jul 24, 2015 #15

    vela

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    How was the logarithm defined in your class? The way I've usually seen it in pre-calc classes is that one says the statement ##\log_b x = u## is equivalent to the statement ##x = b^u##. In this case, you have ##b^{u+v}##, so the log base b of that is, by definition, ##u+v##. There's nothing to prove.
     
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