If $x$ is a real number satisfying the equation $$9\log_3 x - 10\log_9 x =18 \log_{27} 45,$$ then the value of $x$ is equal to $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

[i]Proposed by pog[/i]
1. We start with the given equation:
   \[
   9\log_3 x - 10\log_9 x = 18 \log_{27} 45
   \]

2. We use the change of base formula for logarithms. Recall that:
   \[
   \log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2}
   \]
   and
   \[
   \log_{27} x = \frac{\log_3 x}{\log_3 27} = \frac{\log_3 x}{3}
   \]

3. Substitute these into the original equation:
   \[
   9\log_3 x - 10 \left(\frac{\log_3 x}{2}\right) = 18 \left(\frac{\log_3 45}{3}\right)
   \]

4. Simplify the equation:
   \[
   9\log_3 x - 5\log_3 x = 6\log_3 45
   \]
   \[
   4\log_3 x = 6\log_3 45
   \]

5. Divide both sides by 4:
   \[
   \log_3 x = \frac{6}{4} \log_3 45
   \]
   \[
   \log_3 x = \frac{3}{2} \log_3 45
   \]

6. This implies:
   \[
   x = 45^{\frac{3}{2}}
   \]

7. Simplify \(45^{\frac{3}{2}}\):
   \[
   45^{\frac{3}{2}} = (45^1) \cdot (45^{\frac{1}{2}}) = 45 \cdot \sqrt{45}
   \]

8. Express \(45\) in terms of its prime factors:
   \[
   45 = 3^2 \cdot 5
   \]
   \[
   \sqrt{45} = \sqrt{3^2 \cdot 5} = 3 \sqrt{5}
   \]

9. Therefore:
   \[
   x = 45 \cdot 3 \sqrt{5} = 135 \sqrt{5}
   \]

10. Identify \(m\) and \(n\):
    \[
    m = 135, \quad n = 5
    \]

11. Calculate \(m + n\):
    \[
    m + n = 135 + 5 = 140
    \]

The final answer is \(\boxed{140}\)