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

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. — Sarvesh Ravichandran Iyer, Sep 07 '18 at 16:46
https://math.stackexchange.com/questions/202078/solve-sqrt52-sqrt6x-sqrt5-2-sqrt6x-10 — lab bhattacharjee, Sep 07 '18 at 18:44
Thank you very much for the hints sir @астон вілла олоф мэллбэрг — Luis Szooares, Sep 07 '18 at 22:43
Oh this is similiar with my problem, thank you @lab bhattacharjee — Luis Szooares, Sep 07 '18 at 22:44

score 4 · Answer 1 · edited Sep 07 '18 at 17:02

4

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

edited Sep 07 '18 at 17:02

amWhy

answered Sep 07 '18 at 16:42

Thank you for the advice sir, this is the secon times you have helped me hehe @Dr. Sonnhard Graubner – Luis Szooares Sep 07 '18 at 22:45

1 Answers1