Solving $\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$

Question

Given $$\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$$

What is $x$?

I just can do this with that equation

$$\left(\sqrt{2+1+2\sqrt{2.1}}\right)^x - \left(\sqrt{2+1-2\sqrt{2.1}}\right)^x = \frac32$$

$$\left(\sqrt{({\sqrt{2}+\sqrt{1})^2}}\right)^x-\left(\sqrt{({\sqrt{2}-\sqrt{1})^2}}\right)^x = \frac32$$

$$\left(\sqrt{2}+1\right)^x - \left(\sqrt{2}-1\right)^x = \frac32$$

And i stuck there for a few hours and get nothing

Pliz help me

Answer 1

Use the fact that $$\sqrt{3-2\sqrt{2}}=\frac{1}{\sqrt{3+2\sqrt{2}}}$$