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.

Answer 1

Hint:

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

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

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

Answer 2

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:

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