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B Why does this derivation fail?

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  1. May 28, 2017 #1
    I know that deriving y = bεax gives dy/dx = abεax, I also know that the inverse of dx/dy equals dy/dx, so how come I don't get dy/dx = abεax when I do the following:

    ln(y) = ln(bεax) = ax ln(bε) -> x = ln(y)/(a ln(bε)) -> dx/dy = 1/(ay ln(bε)) -> dy/dx = ay ln(bε) = abεax ln(bε) ≠ abεax

    What am I doing wrong?
     
  2. jcsd
  3. May 28, 2017 #2

    andrewkirk

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    This is where it goes wrong. It should be:
    $$\log y = \log \left(be^{ax}\right) = \log b+ax$$
     
  4. May 28, 2017 #3
    So like this: ln y = ln(bεax) = ln(b) + ax ln(ε) and because ln(ε) = 1 then ln y = ln(b) + ax. Thank you very much
     
  5. Jun 9, 2017 #4

    Mark44

    Staff: Mentor

    The simple answer, and to elaborate on what @andrewkirk said, what you are doing wrong in the first quote above is treating ##be^{ax}## as if it were ##(be)^{ax}## when you take the log. The expressions ##be^{ax}## and ##(be)^{ax}## are different, just as ##2x^3## and ##(2x)^3## are different.

    So ##\ln(be^{ax}) = \ln(b) + \ln(e^{ax}) = \ln(b) + ax##.

    BTW, you should use the letter 'e' for the exponential base, not ε ("epsilon"). At least that's what I think you mean.
     
    Last edited: Jun 9, 2017
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