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.


Question: NTP / SNTP and UTC Time
Answer: Yes, the NTP servers all talk UTC and your local machine translates that into whatever time is appropriate based on your timezone settings.