Answer: 7

Small Hint:

Use $\frac{1}{\log_b a} = \log_a b$ to rewrite both terms as logarithms with base $2000^6$

Big Hint:

The sum becomes $\log_{2000^6}(4^2 \cdot 5^3),$ and $4^2 \cdot 5^3 = 2000$

Solution:

Since $\frac{1}{\log_b a} = \log_a b,$ the two terms equal $2\log_{2000^6} 4 = \log_{2000^6} 16$ and $3\log_{2000^6} 5 = \log_{2000^6} 125.$ Their sum is $$\begin{aligned} \log_{2000^6}(16 \cdot 125) &= \log_{2000^6} 2000 \\ &= \frac{1}{6}. \end{aligned}$$

Since $\gcd(1, 6) = 1,$ the answer is $m + n = 1 + 6 = 7.$