How is the integral calculated in this case? 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 $c \log(x)$
 A: 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 $ln(x)$
