how many possible solution for $\frac{\log a +\log b +\log c}{\log 6}= 6$? How many possible solution for $$\frac{\log a +\log b +\log c}{\log 6}= 6$$
My attempt:
$$\log abc = 6\times \log6$$
$$abc = 6^6$$
Now how can I proceed?
 A: Thee are infinitely many real solutions.
In terms of integer solutions, consider the prime factorization of $6^6=2^63^6$. Now you have a combinatorics problem equivalent to the following:

You have $6$ red balls and $6$ blue balls and three boxes labeled boxes. How many ways are there to put the balls in the boxes?

Notice that the distribution of red and blue balls is independent, and so if $R$ is the number of ways to distribute the red balls and $B$ is the number of ways to distribute the blue balls, the final answer is $RB$. This then reduces the problem again to finding the answer to the following question and squaring it:

How many ways can we distribute $6$ identical balls into $3$ distinguishable boxes

This second problem can be solved using the techniques seen here, which solves the same problem but with different numbers.

We can arrive at this spot using analytic techniques too. Let's look at the original problem and take $\log_{6}$ of both sides, getting $$\log_{6}(a)+\log_{6}(b)+\log_{6}(c)=6$$ How many solutions are there? Well, we can observe that all three of these numbers are integers iff $2$ divides $a$ exactly the same number of times as $3$ does, and same for the other terms. Thus, by the Fundamental Counting Principle, the number of integral solutions $x,y,z$ to the equation $x+y+z=6$ is exactly the square root of the number of integral solutions $a,b,c$ to the equation $\log_{6}(a)+\log_{6}(b)+\log_{6}(c)=6$

The Stars and Bars Theorem is something you should look at for the general case, if you're not familiar with it.
