Why, if the natural logarithm is an increasing function, then we can use it for finding the same maximum of $L(p)$?

Question

This might seem a trivial problem, but I would like to be sure about my intuitions.

In the method of maximum likelihood estimation, people use very often the natural logarithm in order to simplify the differentiation of the function; specifically, I was reading the following:

That is, if $x1 < x2$, then $f(x1) < f(x2)$. That means that the value of $p$ that maximizes the natural logarithm of the likelihood function $ln(L(p))$ is also the value of $p$ that maximizes the likelihood function $L(p)$.

My question is

Why, if the natural logarithm is an increasing function, then we can use it for finding the same maximum of $L(p)$?

I understand that the differentiation of $L(p)$ will be easier, but I am not seeing why the critical points would be the same, and thus the maxima would be the same.

Answer 1

Suppose $\ln(L(p))$ is maximal for $p=p^*$, but it does not maximize $L(p)$. Then, there exists $\tilde{p}\neq p^*$ such that $L(\tilde{p})>L(p^*)$. Because the logarithm is increasing, this also means that $\ln(L(\tilde{p}))>\ln(L(p^*))$. This is a contradiction with $\ln(L(p^*))$ being maximal.

Therefore, if you find the maximum of $\ln(L(p))$, this gives you the maximum of $L(p)$.