It's a perfect square

$\sqrt{3+\sqrt2}+\sqrt{3-2\sqrt2} \\ =\sqrt{(\sqrt2+1)^2}+\sqrt{(\sqrt2-1)^2}=\sqrt2+1+\sqrt2-1\\ =2\sqrt2 \\ \therefore \color{#20A900}{a}=\color{#20A900}{2\sqrt2}$ Now, $\log_{\color{#20A900}{2\sqrt2}}{2^9}=9\log_{\color{#20A900}{2^{\frac{3}{2}}}}{2} \\ =\dfrac{9\cdot2}{3}\log_{2}{2} \\ =\boxed{6}.$

Note : $\large{\log_{a}{b^c}=c\log_{a}{b}}$ and $\large{\log_{a^c}{b}=\dfrac{1}{c}\log_{a}{b}}.$

Since,
$\large \begin{aligned} \sqrt{3+2\sqrt{2}} + \sqrt{3- 2\sqrt{2}} = \sqrt{(\sqrt{2} + 1)^2} +\sqrt{(\sqrt{2} - 1)^2} \\ = \sqrt{2} + 1 + \sqrt{2} - 1 = \color{#3D99F6}{2\sqrt{2}}\end{aligned}$

$\large \log_{\color{#3D99F6}{a}} 2^9 = \log_{2\sqrt{2}} 2^9 = \dfrac{9}{3/2}\log_2 2 = \boxed{6}$

Note : $\large \log_{a^b} c^d = \dfrac{d}{b} \log_a c \quad ; \quad \log_a a = 1$