Quickie: say you have the expression;

$$\frac{d\ln(x)}{d\ln(y)} = z$$

Is it legit to raise both sides by $e$ so it becomes

original question

$$\frac{dx}{dy} = e^z$$

edited because I forgot about math

$$ e^{\left( \dfrac{d \ln x}{d \ln y}\right)} = e^z$$

  • $\begingroup$ What is your definition of the expression $d\ln x/d\ln y$? $\endgroup$ – Stahl Mar 21 '13 at 22:08
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    $\begingroup$ It is legit to raise both sides by $e$ so that it becomes $e^ { \frac { d\, \ln x} { d \, \ln y} } = e^z$. $\endgroup$ – Calvin Lin Mar 21 '13 at 22:09

What you wrote is pure formality, I've never seen such a notation. But if want to follow all rules of differentiation and etc., you could have done this $$ d \ln x = \frac {dx}x \\ d \ln y = \frac {dy}y \\ \frac {d \ln x}{d \ln y} = \frac {y\,dx}{x\,dy} = z \\ \frac {dx}{dy} = z\frac xy $$

  • $\begingroup$ OK so this is cool. Not to sound too stupid (though I think I blew that in my original post) but why is $d\ln x = \dfrac{dx}{x}$? $\endgroup$ – Alex Mar 21 '13 at 22:18
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    $\begingroup$ Denote $f = \ln x$, so its derivative $f' = \frac {df}{dx} = \frac 1x$, from which you can find $df = \frac {dx}x$, and just recall that $df = d\ln x$. just like Sami put in his answer. $\endgroup$ – Kaster Mar 21 '13 at 22:20
  • $\begingroup$ Now that you've written it it seems trivial, but this is a major "penny drop" moment for me. Thank you. $\endgroup$ – Alex Mar 21 '13 at 22:54
  • $\begingroup$ @Alex no problem. $\endgroup$ – Kaster Mar 21 '13 at 23:54

We have $df(x)=f'(x)dx$ so $$z=\frac{d\ln x}{d\ln y}=\frac{y\,dx}{x\,dy}$$ hence $$\frac{dx}{dy}=z\frac{x}{y}.$$


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