Logs are complicated...

Exp. = 2./.(6log(2000)./.log(4)) + 3./.(6log(2000)./.log(5))
=(2log(4) )./.(6log(2000) + (3log(5) )./.(6log(2000)
=(log(16) + log(125)),/,(6log(2000) = log(2000)./.(6log(2000) =1/6 =a/b.

The number can be rewritten as $\frac{2}{6\log_4{2000}}+\frac{3}{6\log_5{2000}}$ . Then, we can reverse the logarithms to make $\frac{2}{\frac{6}{\log_{2000}{4}}}+\frac{3}{\frac{6}{\log_{2000}{5}}}$ . We can then rewrite again to get $\frac{\log_{2000}{16}+\log_{2000}{125}}{6}$ . This equates to $\frac{\log_{2006}{2006}}{6}$ , which is (trivially) equal to $\frac{1}{6}$ . Thus we have 1+6 which equals $\boxed{7}$ .