I'm stuck on continuing the next exercise:
Considering:
$\log_{c}a = 3$
$\log_{c}b = 4$and:
$ y = \frac{a^{3}\sqrt{b \cdot c^{2}}}{2} $
What's the value of $\log_{c}y$ (integer)?
So far, I did all the substitutions that were obvious at my eyes:
$ y = \frac{a^{3}\sqrt{b \cdot c^{2}}}{2}\quad\rightarrow\quad y = \frac{c^{9}\sqrt{c^{4} \cdot c^{2}}}{2}\quad\rightarrow\quad y = \frac{c^{9}\sqrt{c^{6}}}{2}\quad\rightarrow\quad y = \frac{c^{9} c^{3}}{2}\quad\rightarrow\quad y = \frac{c^{12}}{2} $
The last equality is the same as $c^{12} = 2y$, which could be written as:
$ \log_{c}2y = 12\quad\rightarrow\quad \log_{c}2 + \log_{c}y = 12 $
I really don't know haw to continue. I've managed to find a $\log_{c}y$, but I don't know what to do with the $\log_{c}2$. Either I took the wrong path, or I'm missing something that prevents me to finish this exercise.
Any hints are much welcome! Thanks in advance.