Denesting Logarithmic expressions $\log_7(\log_2(3)) + \log_7(\log_5(6)) + \log_7(\log_{11}(1/2)) = \log_7(-1) + \log_7(\log_5(3)) + \log_7(\log_{11}(6))$
This can only be simplified by using the sum to product rule and noticing that of the $n!$ possible permutations that $\dfrac{1}{2}$ is a power of $2$. Because N is small and the relationship between $2$ to $\dfrac{1}{2}$ is simple this can be solved trivially. How does one go about simplifying much more complex logarithmic identities in polynomial time?
 A: If I'm understanding what you're looking for (and that $-1$ in parentheses on the right-hand side is illegal, as we'll come to below), your expression works because of the "Change-of-Base" Formula, $\log_b (x) = \frac{\log_a (x)}{\log_a (b)}$.
So, as you say, because the sum of logarithms gives the logarithm of a product, you can carry on permutations of the sort you have here for any number of terms:
$$\log_a [ \log_{b_1}(x_1) ] + \log_a [ \log_{b_2}(x_2) ] + \ldots + \log_a [ \log_{b_n}(x_n) ]$$
$$=  \log_a [ \log_{b_1}(x_1) \cdot \log_{b_2}(x_2) \cdot  \ldots  \cdot  \log_{b_n}(x_n) ]$$
$$=  \log_a [ \frac{\ln (x_1)}{\ln (b_1)} \cdot \frac{\ln (x_2)}{\ln (b_2)} \cdot  \ldots  \cdot  \frac{\ln (x_n)}{\ln (b_n)} ] = \log_a [ \frac{\ln (x_1) \cdot \ln (x_2) \cdot \ldots \ln (x_n)}{\ln (b_1) \cdot \ln (b_2) \cdot \ldots \ln (b_n)  }] . $$
You can see from this that you are then free to permute the factors in the product in either the numerator or the denominator to create any combination of numbers $x_i$ and logarithmic bases $b_i$ you wish.
OK, having said that, I see what is going on in your equation. There is a domain issue here in order to apply this result.  The nested value $\log_a(\log_b(x))$ must be defined for every term in the sum. Thus, you cannot have any terms where any of the $x_i$ are less than or equal to 1; otherwise, you will have a term with $\log_{b_i}(x) \le 0$, for which you will not be able to evaluate $\log_a(\log_{b_i}(x))$.
