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

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

• Note that $\sqrt 2 - 1 = \frac 1{\sqrt 2 + 1}$. This will allow you to set $(\sqrt 2 +1)^x = y$ , then solve for $y$ as a quadratic equation then retrieve $x$ from the value(s) of $y$ obtained. – Teresa Lisbon Sep 7 '18 at 16:46
• math.stackexchange.com/questions/202078/… – lab bhattacharjee Sep 7 '18 at 18:44
• Thank you very much for the hints sir @астон вілла олоф мэллбэрг – Luis Szooares Sep 7 '18 at 22:43
• Oh this is similiar with my problem, thank you @lab bhattacharjee – Luis Szooares Sep 7 '18 at 22:44

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