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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

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    $\begingroup$ 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. $\endgroup$ – Teresa Lisbon Sep 7 '18 at 16:46
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    $\begingroup$ math.stackexchange.com/questions/202078/… $\endgroup$ – lab bhattacharjee Sep 7 '18 at 18:44
  • $\begingroup$ Thank you very much for the hints sir @астон вілла олоф мэллбэрг $\endgroup$ – Luis Szooares Sep 7 '18 at 22:43
  • $\begingroup$ Oh this is similiar with my problem, thank you @lab bhattacharjee $\endgroup$ – Luis Szooares Sep 7 '18 at 22:44
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Use the fact that $$\sqrt{3-2\sqrt{2}}=\frac{1}{\sqrt{3+2\sqrt{2}}}$$

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  • $\begingroup$ Thank you for the advice sir, this is the secon times you have helped me hehe @Dr. Sonnhard Graubner $\endgroup$ – Luis Szooares Sep 7 '18 at 22:45

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