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

Solution 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!

Solution 2:

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.

Solution 3:

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

Solution 4:

Kind of intuitive answer: Maximising $\ln f$ involves taking the derivative: $\frac{d \ln f(x)}{dx}$ and setting it equal to zero, and maximising $f$ involves taking the derivative: $\frac{d f(x)}{dx}$ and setting it equal to zero.

$$\frac{d \ln f(x)}{d x} = \frac{f'(x)}{f(x)}$$

Thus

$$\frac{d \ln f(x)}{d x} = 0 \to \frac{f'(x)}{f(x)} = 0 \to f'(x) = 0$$

Update: I prove for the $\ln$ case instead of the general case of increasing function $g$:

Maximizing log(x)