Q:

How to write Kelvin equation in a different way

My question is about the Kelvin equation which is as follows:
$$\ln(e/e_s) = \frac{2\cdot\sigma}{n\cdot k\cdot T\cdot r}  $$
Keep in mind that the $e$ in $\ln(e/e_s)$ is not the constant $\mathrm{e}$.
I know that: 
$$\frac{e}{e_s} = \mathrm{e}^{(\frac{2\cdot\sigma}{n\cdot k \cdot T \cdot r})}$$
I need to show that this is equal to:
$$\frac{e}{e_s} = 1 + \frac{a}{r}$$
Where $a = \frac{2\cdot \sigma}{n\cdot k\cdot T}$
How do I go about doing this?

A:

You want to show that $e^{a/r} = 1 + a/r$.  But that isn't true.  What is true is that
$$ e^{a/r} = 1 + a/r + \frac{a^2}{2!r^2} + \frac{a^3}{3!r^3} + \cdots$$
So if $a/r$ were small, so that $a^2/r^2$ and all the following terms on the right were very small, then $1+a/r$ would be a good approximation for $e^{a/r}$.

