I have some issues with using the logarithm formulas, I have this expression for example where $\log$ is the natural logarithm:
$$ \log ( \frac{1}{3}\theta^{3y}) $$
Then we know the standard logarithm rules: $$ \log(a^b) = b\log(a) $$ $$ \log(\frac{a}{b}) = \log(a)-\log(b) $$
But If I apply them in various order I get two different results: $$ \log(\theta^{3y}) - \log(3) = 3y \log(\theta) - \log(3) $$
I am working through an old exam set, and the following is what my professor got in the same calculation: $$ \log ( \frac{1}{3}\theta^{3y}) = 3y\log(\frac{\theta}{3}) = 3y(\log(\theta)-\log(3)) = 3y\log(\theta) - 3y\log(3) $$
I think that my mistake might be that I am disregarding that exponents binds tigther than division so I can only apply these logarithm rules to the operator that binds the tightest at that moment? so my question is about composite functions and how to reason about when I can use which rules?
Except for the last part of your question, all the logarithm rules applied are correct . For instance,
$$\log \left( \frac{1}{3}\theta^{3y}\right) = 3y\log\left(\frac{\theta}{3}\right)$$
This is not correct. Because, observe that :
$$ \begin{align}3y\log\left(\frac{\theta}{3}\right)&=\log\left(\left(\frac{\theta}{3}\right)^{3y}\right)\\ &=\log\left(\frac{\theta}{3}\right)^{3y}\\ &=\log \left( \frac{\theta^{3y}}{3^{3y}}\right)\\ &≠\log\left(\frac {\theta^{3y}}{3}\right)\\ &=\log \left( \frac{1}{3}\theta^{3y}\right)\end{align} $$
or
$$ \begin{align}3y\log\frac{\theta}{3}&=\log\left(\frac{\theta}{3}\right)^{3y}\\ &=\log \frac{\theta^{3y}}{3^{3y}}\\ &≠\log\frac {\theta^{3y}}{3}\\ &=\log\left(\frac{1}{3}\theta^{3y}\right)\tag {*}\end{align} $$
The right steps can be written as follows :
$$ \begin{align}\log \left( \frac{1}{3}\theta^{3y}\right)&=\log\frac 13+\log \left(\theta^{3y}\right)\\ &=-\log 3+3y\log \theta\end{align} $$
Finally, note that the logarithm rules are hold within certain restrictions or in the domain in which they are defined :
$$\log \left(\theta^{3y}\right)=3y\log \theta$$
holds iff, when $\theta >0$ and $y\in\mathbb R\thinspace .$
$(^*)$
$\rm {Small \thinspace \thinspace supplement :}$
The following notations are exactly the same :
$\log(a^b) = \log a^b$
$\log\left(\dfrac{a}{b}\right )=\log \dfrac ab$
But,
$$3y\log\left(\frac{\theta}{3}\right)=\log\frac{{\theta}^{3y}}{{3}}$$
is not correct . The correct notation is :
$$3y\log\left(\frac{\theta}{3}\right)=\log\left(\frac{\theta}{3}\right)^{3y}$$
Because, when writing notation, not using parentheses where necessary leads to different results :
$$ \begin{align}\left(\frac {\theta}{ 3}\right)^{3y}=\frac {\theta^{3y}}{3^{3y}}≠{\frac {\theta} {3}}^{3y}=\frac {\theta^{3y}}{3}\thinspace .\end{align} $$