Hi I am an IB student and am stuck on part of a problem I am trying to solve. The problem is in regard to integrating a function of the form $y=\frac{c}{x}$.

How does the author proceed from:

$$\left(\frac{c}{x}-d\right)\dot{x}+\left(\frac{a}{y}-b\right)\dot{y}=0 $$ to $$\frac{d}{dt}[c\log(x)-dx+a\log(y)-by]=0 $$

Isnt the integral of:

$\frac{c}{x}=c\ln(x)$ and not $c \log(x)$

  • 5
    $\begingroup$ log base 10 was seldom used except in numerical calculations (and not today since we an use calculators). Generally, in college Calculus or higher math classes, "log" is used to mean the natural logarithm rather than "ln". $\endgroup$
    – user247327
    Dec 28 '17 at 18:42
  • $\begingroup$ @user247327 oh so in this case log is not base 10 but base e? $\endgroup$ Dec 28 '17 at 19:07
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – user247327
    Dec 28 '17 at 19:39

Quoting a user from Reddit: https://www.reddit.com/r/learnmath/comments/1upa40/calculus_why_is_the_integral_of_1x_lnx_and_not/

"Some mathematicians will use "log(x)" and "ln(x)" interchangeably. Others will take "log(x)" to mean log base 10, and "ln(x)" to mean log base e. I belong to the latter group.

With that said, the integral of $\frac{1}{x}$ can be thought of as $ln(x)$ either by definition or by understanding that the derivative of $ln(x)$ is $\frac{1}{x}$.

There are several neat proofs to show you that the derivative of $ln(x)$ is $\frac{1}{x}$, but I think the easiest one is the one qoppaphi showed using implicit differentiation.

$y = ln(x)$

re-write in exponential form

$e^y = x$

take derivative with respect to x

$e^y * \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{e^y}$

$\frac{dy}{dx} = \frac{1}{x}$

This proof requires you to know that the derivative of $e^u$ is $e^u * \frac{du}{dx}$"

To answer your question specifically... you're both correct, the given is simply using $log(x)$ to refer to $ln(x)$

  • $\begingroup$ Ahh that makes sense :) Thank you! $\endgroup$ Dec 28 '17 at 19:14

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