Why $-\ln|\cos x| = \ln|\sec x|$

Question

I'm currently learning derivatives and antiderivatives. In a solution to one of the question, it mentions

$$-\ln|\cos x| +C=\ln|\cos x|^{-1}+C= \ln|\sec x|+C$$

I am puzzled by what rules this transformation uses. Could someone please kindly point it out to me.

Edit 1:

I think my main confusion is the use of absolute value and natural log in one place.

Answer 1

You should read $\ln |\cos x|^{-1}$ as

$$\ln(|\cos x|^{-1})$$

and $-\ln |\cos x|$ as

$$(-1) \ln(|\cos x|)$$

Now remember the rules that $a \ln(b) = \ln(b^a)$ and $\frac{1}{\cos x} = \sec x$.

The absolute value does not affect the log rule - it's inside the log, and so effectively stands for a number, albeit variable, being passed into it.

Answer 2

$$\sec x :=\frac 1 {\cos x} \implies \log|\sec x|=-\log|\cos x|$$

Answer 3

$\sec x = \frac{1}{\cos x} = \cos x^{-1}$

Hence

$\ln|\sec x| = \ln |\cos x^{-1}| = (-1)(\ln|\cos x|)=-\ln|\cos x|$