TITLE: Homework with logarithms
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: Note that we know $\log_c y$ is an integer, so $\log_c 2$ must also be an integer, hence $c^k=2$ for some $k\in\mathbb{Z}$.  If we also assume $c$ is an integer, then the problem does in fact have a unique solution.  Hope this helps.