Differential equation of first order

Question

I have this simple differential equation: $y'=(\tan x)y.$

after integrating $\frac{y'}{y(x)}= \tan x$

i came up with $\log y(x)=-\log \cos(x)+1.$

now my question is this one: why $ e^{-\log \cos(x)}=\sec(x)?$

thank you for your time

Answer 1

I suppose that $\log(x)$ is for $\ln(x)$ $$e^{-\log \cos(x)}=(e^{\log \cos(x)})^{-1}=(\cos(x))^{-1}=\frac 1 {\cos(x)}=\sec(x)$$