What happen to units when we take Ln? What happens to units when we take the Ln of the both side of the equation?
For example for an equation $PV^k = C$ (which becomes $\ln P = -k \ln V + \ln C$), how can we find the units of $k$ and $C$?
-->(the unit of $P$ is mmHg and the unit of $V$ is $cm^3$)
Thank you so much..
 A: In the sciences we never take the log of a number with a unit.  For any unitless $k$, $\ln(k)$ is defined. But for a $k$ with a unit, $\ln(k)$ is typically avoided. Its undefined. Its undefined because the answer you get and the relationships you find end up being dependent on the unit, and not on the pure mathematics. Strictly speaking this doesnt really matter. Every equation - in the sciences at least - use constants and ratios to convert units, but the idea is to avoid the need.
The same logic holds with trig functions. The evaluation of $\sin(x)$ will change depending on if youre using radians, grads, degrees, or any other unit.  The radian is a formal standard, and there are a number of mathematical reasons for that.
Often what we do is have a constant that solves the problem. $\ln(kc)$, where $c$ is the inverse units of whatever units $k$ has. Thus $kc$ becomes unitless.  Sometimes that $c$ constant is hidden and presumed, and youre simply told "make sure youre using mks units". Similarly with trig functions, $\sin(kx)$ works for degrees if $k$ is in units radians per degree.

As a matter of fact, we can take this a step further. Beyond the scope...
Inside of trig functions such as $\sin(x)$, $x$ is treated as "unitless" for the same argument. But in fact it does have a unit, the radian.  This is the scientific and mathematical standard.  A radian is sort of a unitless unit; an angle that represents the ratio between the length of an arc to the length of a radius, $\frac{\mathrm{length}}{\mathrm{length}}$, what you might think is unitless is in context just a radian. What comes out of the trig function though is still unitless.
We can do a similar thing with exponentials.  The unit Neper, $\mathrm{Np}$, is the unit on the value $x$ that you stick into an exponential $e^x$ to get a unitless solution.  Naturally it kind of makes sense for the values inside the argument and outside the function to be different. The Neper is a sort of unitless unit that represents the logarithm of a unitless number.
A: It should be obvious that in the $PV^k=C$ equation the value of $C$ depends on the units of $P$ and $V$. For example, one can measure volume $V$ is in $m^3$, or liters, or $ft^3$. Then I would write the above equation as $$\frac P{Nm^{-2}}\frac{V^k}{m^{3k}}=\frac{C}{Nm^{3k-2}}$$
Then one can use dimensionless quantities $p=\frac P{Nm^{-2}}$, $v=\frac v{m^3}$, $c=\frac{C}{Nm^{3k-2}}$.
With this, you can take the logarithm
$$\ln p=-k\ln v+\ln c$$
A: Well, $k$ must be unitless, because exponentiation isn't defined for anything except numbers.
So the units of $C$ are $mmHg * cm^{3k}$.
