Log Mania!

To expand just a bit on what Aaghaz already stated:

For convenience, can rewrite the equation as

$\dfrac{2\log{a}}{3\log9} + \dfrac{3\log9}{5\log{c}} + \dfrac{5\log{c}}{2\log{a}}=3$

Then, since logs are strictly positive, we can apply AM-GM to the left side, giving

$\begin{aligned} \dfrac{\dfrac{2\log{a}}{3\log9} + \dfrac{3\log9}{5\log{c}} + \dfrac{5\log{c}}{2\log{a}}}{3} &\ge \sqrt[3]{\dfrac{2\log{a}}{3\log9} \cdot \dfrac{3\log9}{5\log{c}} \cdot \dfrac{5\log{c}}{2\log{a}}} = 1 \\ \\ \dfrac{2\log{a}}{3\log9} + \dfrac{3\log9}{5\log{c}} + \dfrac{5\log{c}}{2\log{a}} &\ge 3 \end{aligned}$

Also by AM-GM, equality in the above statements occurs if and only if the terms are themselves all equal, which then gives

$\begin{aligned} \dfrac{2\log{a}}{3\log9} &= 1 \\ 2\log{a} &= 3\log9 \\ a^2 &= 9^3 \\ a &= 3^3 = \boxed{27} \\ \end{aligned}$

(We can discard the negative solution to the above quadratic as both arguments and bases of logs must be positive.)

A simple application of AM-GM.....