Why does finding the $x$ that maximizes $\ln(f(x))$ is the same as finding the $x$ that maximizes $f(x)$?

Question

I'm reading about maximum likelihood here.

In the last paragraph of the first page, it says:

Why does the value of $p$ that maximizes $\log L(p;3)$ is the same $p$ that maximizes $L(p;3)$. The fact that it mentions that log is an increasing function does not help me understand it at all. Does this method work for any increasing function?

I've covered single and two variable optimization but have not come across this fact until now. I'd like to understand why it must be true and read more about its explanation. So, I would also appreciate it if someone can provide some links to it (I don't know what term to search).

Answer 1

Yes, this works for any increasing function. Consider the following: Let $f(P)$ be some real-valued quantity which you want to maximize, that depends on a family of parameters $P$.

Also let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly monotonously increasing function, that is $t>s$ is equivalent to $g(t)>g(s)$.

Suppose you have values $P_0$ for the parameters, such that $f(P_0)$ is a strict maximum. That means we have $f(P_0)>f(P)$ for all parameter values $P\ne P_0$. This is by the above equivalent to $$g(f(P_0))>g(f(P))$$ setting $t=f(P_0)$ and $s=f(P)$, which exactly means that $g(f(P_0))$ is maximal!

Answer 2

Because the function $\ln$ is increasing. More details?

Answer 3

The $p$ that maximizes $L(p;3)$ (if it exists) is some $p^*$ so that $L(p^*;3)\ge L(p;3)$ for all $p$. That is what maximizing means.

Since $\log$ is increasing, $L(p^*;3)\ge L(p;3)$ implies $\log L(p^*;3)\ge \log L(p;3)$. So $p^*$ maximizes $\log L(p;3)$ as well.