Answer: 54

Small Hint:

Set $x = \log_a b;$ since $\log_b a = \frac{1}{x},$ the equation becomes $x + \frac{6}{x} = 5$

Big Hint:

The two cases are $b = a^2$ and $b = a^3;$ count how many $a \ge 2$ keep $b \le 2005$ in each

Solution:

Let $x = \log_a b.$ Since $\log_b a = \frac{1}{x},$ the equation becomes $x + \frac{6}{x} = 5,$ i.e. $x^2 - 5x + 6 = 0,$ so $x = 2$ or $x = 3.$ That means $b = a^2$ or $b = a^3.$

For $b = a^2 \le 2005$ we need $2 \le a \le 44$ (since $44^2 = 1936$ and $45^2 = 2025$), giving $43$ pairs. For $b = a^3 \le 2005$ we need $2 \le a \le 12$ (since $12^3 = 1728$ and $13^3 = 2197$), giving $11$ pairs.

In total there are $43 + 11 = 54$ ordered pairs.