Determine the number of [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(a,b)$ of [integers](https://artofproblemsolving.com/wiki/index.php/Integer) such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$
The equation can be rewritten as $\frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a  \log b}=5$ Multiplying through by $\log a \log b$ and factoring yields $(\log b - 3\log a)(\log b - 2\log a)=0$. Therefore, $\log b=3\log a$ or $\log b=2\log a$, so either $b=a^3$ or $b=a^2$. 

For the case $b=a^2$, note that $44^2=1936$ and $45^2=2025$. Thus, all values of $a$ from $2$ to $44$ will work.
For the case $b=a^3$, note that $12^3=1728$ while $13^3=2197$. Therefore, for this case, all values of $a$ from $2$ to $12$ work.
There are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= \boxed{054}$.
Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs $(a,b)$, and not for the number of possible values of $b$. Were the problem to ask for the number of possible values of $b$, the values of $b^6$ under $2005$ would have to be subtracted, which would just be $2$ values:  $2^6$ and $3^6$. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)