Exponential equation: $(3-2\sqrt2)^x + (3+2\sqrt2)^x = 6$

Question

Equation: $(3-2\sqrt2)^x + (3+2\sqrt2)^x = 6$

I tried using logarithms and typical operations. Please advice.

You can potentially find all solutions for $x\in\mathbb N$ by assuming $x$ is a natural number and binomially expanding. — Mark Schultz-Wu, Dec 13 '16 at 17:13

score 6 · Accepted Answer · answered Dec 13 '16 at 17:12

6

Hint:

$$3-2\sqrt{2} = \dfrac{1}{3+2\sqrt{2}}$$

Now let $(3+2\sqrt{2})^x = t$. We have

$$t+\dfrac{1}{t} = 6$$

answered Dec 13 '16 at 17:12

Max Payne

Arnaldo · Answer 2 · 2016-12-13T17:49:40.410

5

Hint

Call $p=(3-2\sqrt2)^x$ and $q=(3+2\sqrt2)^x$. Note that:

$p+q=6$ and $p\cdot q=1$. So you will get:

$$q^2-6q+1=0$$

and solving that you get:

$$q \in \{3+2\sqrt{2},3-2\sqrt{2}\}$$

edited Dec 13 '16 at 17:49

answered Dec 13 '16 at 17:12

Arnaldo

Or $q=3-2\sqrt{2}$ of course... So $x=-1$ is also a solution. – WimC Dec 13 '16 at 17:47
@Wim, of course! – Arnaldo Dec 13 '16 at 17:48

2 Answers2