Prove that the likelihood function L(θ|x) is equivalent to maximizing log L(θ|x) where log is the natural logarithm In other words, why 
$\text{argmax} \text{ } L(\theta) = \text{argmax} \text{ } \text{log} \text{ } L(\theta)$
?
 A: Additionnally to @forgottenscience answer, let consider the following (more intuitive, to my view) point:
First,

*

*$x \rightarrow \log(x)$ is a strictly increasing function


*Let call $\theta^*$ the value maximizing $L$ and $\nu^*$ the value maximizing $\log(L)$
(i.e. $\theta^* =\arg\max_{\theta} L(\theta)$ and $\nu^* =\arg\max_{\nu} \log L(\nu)$).
Then, if $\nu^* \ne \theta^*$

*

*then necessarily $L(\theta^*) > L(\nu^*)$ (remember that by definition $\theta^*$ is the value maximizing $L$),


*then necessarily $\log(L(\theta^*)) > \log(L(\nu^*))$ (remember  $\log$ is a strictly increasing function)


*which is impossible as, by definition, $\nu^*$ is the value that makes $\log(L(\nu))$ maximal
As a conclusion, $\nu^* == \theta^*$ i.e. $\arg\max L(\theta)=\arg\max \log L(\theta)$
And this is actually true for any other function that is strictly increasing, which may be convenient in many situation.
A: I'll assume that the likelihood is differentiable and has a unique maximum. If $\theta^*$ is the argmax of $L(\theta)$, 
$$\frac{\partial}{\partial \theta} L(\theta^*) = 0, $$
while it follows from the chain rule that
$$\frac{\partial}{\partial \theta} \log L(\theta) = \frac{\frac{\partial}{\partial \theta} L(\theta)}{L(\theta)}, $$
so in particular
$$ \frac{\partial}{\partial \theta} \log L(\theta^*) = \frac{\frac{\partial}{\partial \theta} L(\theta^*)}{L(\theta^*)} = 0,$$
showing that $\theta^*$ also is the argmax of $\log L(\theta)$.
A: This is true for any strictly increasing function, and it is quite trivial to prove using the definitions of strict monotonicity and the argmax.  In fact, the result does not ]require either of the functions to be differentiable.  If $f$ is any strictly increasing real function then, by definition, we have:
$$L > L'
\quad \quad \iff \quad \quad
 f(L) > f(L').$$
It follows that:
$$L \geqslant L'
\quad \quad \iff\quad \quad
 f(L) \geqslant f(L').$$
Thus, for any function $L:\Theta \rightarrow \mathbb{R}$, we have:
$$\begin{aligned}
\underset{\theta \in \Theta}{\text{arg max }} f(L(\theta))
&= \{ \theta \in \Theta | (\forall \theta' \in \Theta): f(L(\theta)) \geqslant f(L(\theta')) \} \\[6pt]
&= \{ \theta \in \Theta | (\forall \theta' \in \Theta): L(\theta) \geqslant L(\theta') \} \\[12pt]
&= \underset{\theta \in \Theta}{\text{arg max }} L(\theta), \\[6pt]
\end{aligned}$$
where the second step follows from the strict monotonicity of $f$ (in particular, the fact that $f(L) \geqslant f(L')$ implies $L \geqslant L'$).  Note also that for any non-decreasing function $f$ the same general process gives the weaker result that $\underset{\theta \in \Theta}{\text{arg max }} f(L(\theta)) \supseteq \underset{\theta \in \Theta}{\text{arg max }} L(\theta)$.
