Exponential approximate result How come:
$e^{M\ln(1+\mu \delta t)} \approx e^{\mu M \delta t}$ ?
This implies that:
$M\ln(1+\mu \delta t) \approx \mu M \delta t$
But I can't see how this is the case
I think this may be a bit out of context. I just realized. Mu is the mean return on the stock, delta t is time step and M is the number of time steps. Now that I have mentioned this, I think that the question is perhaps more appropriate in QF.
EDIT: I suppose this is not a question of precalculus algebra then.
 A: $$\ln(1+x) \sim x$$
In fact, for $\vert x \vert <1$, we have
$$\ln(1+x)= \sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^k}k$$
A: 
I thought it would be instructive to show two approaches, which have not yet been posted as answers, that establish the approximation in question.  The first relies only on standard tools (not calculus) while the second is based on the integral definition of the logarithm function.


First, in THIS ANSWER, I used only elementary, non-calculus based tools that the logarithm function $\log x$ satisfies the inequalities
$$\frac{x}{1+x}\le \log (1+x) \le x \tag 1$$

Alternatively, we can use the integral definition of the logarithm function to establish the same bounds.  That is, 
$$\log (1+x) =\int_1^{1+x} \frac{1}{u}\,du \implies \frac{x}{1+x}\le \log x \le x$$

Therefore, we have from $(1)$
$$M\frac{\mu dt}{1+\mu dt} \le M\log (1+\mu dt)\le M\mu dt \tag 2$$
Note that the right-hand side of $(2)$ is the approximation of interest; this is an upper bound.  The left-hand side can easily be shown to be greater than $M\mu dt-M(\mu dt)^2$.  And so we have
$$M\mu dt(1-\mu dt) \le M\log (1+\mu dt)\le M\mu dt$$
