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Definition of e / Changing base question

  1. Nov 25, 2014 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Here's the problem:

    53d7bdc8-acbf-4a00-8d2c-05628386088b_zps73e78890.png
    Mod note: the last line is extraneous, and unrelated to this problem.

    2. Relevant equations

    Definition of e (1-1/n)^n, etc.

    3. The attempt at a solution

    My question is this:
    a) When do you know to use e in a situation like this? Is it when you have exponents in such a situation?
    b) Is this the definition of e? Or utilizing it somehow?
    c) If there is any sites/information that you can give me on this procedure, I would be grateful. In my calc studies this must have been skipped over and now it is being applied as if it has been taught to me already.
     
    Last edited by a moderator: Nov 25, 2014
  2. jcsd
  3. Nov 25, 2014 #2

    Mark44

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    Where did the last line come from? The value of this limit is zero, and is totally unrelated to the problem of finding the derivative of ##2^{x + 1}##.
    An exponential function using an arbitrary base can be rewritten as an exponential function use e as the base. The basic idea is that ##a = e^{ln(a)}##, so ##a^x = (e^{ln(a)})^x = e^{xln(a)}##
     
  4. Nov 25, 2014 #3

    RJLiberator

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    Mark44, sorry for the confusion, please exclude the last line. This was a snapshot of a calculation for a longer problem. I only had a vague idea of what was happening in this part and was looking for clarification :).

    Thank you for your definition. I had that idea in my mind, but needed it laid out in front of me as you did. This helped me understand when to use e in a limit problem such as above. :)

    Cheers
     
  5. Nov 25, 2014 #4

    Mark44

    Staff: Mentor

    But the problem you posted is not a limit problem (unless you invoke the underlying definition of the derivative). I think it would be clearer to describe it as a derivative problem.
     
  6. Nov 25, 2014 #5

    RJLiberator

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    Ah yes, this was a limit problem applying l'hospitals rule so it became a derivative problem inside a limit problem. :)

    Indeed, it should have been labeled as a derivative problem according to my issue.
     
  7. Nov 25, 2014 #6

    Ray Vickson

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    You wrote
    [tex] 2^{x+1} = e^{\ln(2)(x+1)}\\
    = \frac{d}{dx} \left( e^{\ln(2)(x+1)} \right) [/tex]
    Do you see why this statement is wrong?
     
  8. Nov 25, 2014 #7

    Fredrik

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    I think the equality on the second line was meant to continue the calculation on the line before the first line. The grayed out line is only the explanation for why the equality holds.
     
  9. Nov 25, 2014 #8

    Fredrik

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    a) You're supposed to prove that ##\frac{d}{dx}2^{x+1}=2^{x+1}\ln x## for all x. If you're allowed to use the formula ##\frac{d}{dx}a^x =a^x\log_a x##, then the proof is a trivial application of that formula and the chain rule. If you're not allowed to use it, then you have to rewrite ##2^{x+1}## as something that you know how to deal with, and apparently you know how to deal with ##e^{f(x)}## when f is a differentiable function, so you write ##2^x=e^{\ln 2^x}=e^{x\ln 2}##.

    b) If your book defined the exponential function before the logarithm, then it probably defined the logarithm as the inverse of the exponential function. So ##e^{\ln x}=x## and ##\ln e^x=x## follow immediately from the definition of the logarithm.

    c) The ##x=e^{\ln x}## rewrite is very common, so it should be mentioned in any book on calculus. It's basically just the definition of the logarithm, so there isn't much to say about it, but you should be able to find examples of high to use it in any book on calculus that includes some solved exercises.
     
  10. Nov 25, 2014 #9

    RJLiberator

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    Excellent material Fredrik. This will help me out, surely.
     
  11. Nov 25, 2014 #10

    Ray Vickson

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    Yes, I know that, but I was hoping the OP would see where the writing error lies, and promise to do better in future.
     
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