How to solve this equation involving $()^x$? I have the equation:
$\left (\sqrt{3+2\sqrt{2}} \right )^x- \left (\sqrt{3-2\sqrt{2}} \right )^x=\frac{3}{2}$
I wrote the left side of the equation as square roots.
$(1+\sqrt{2})^x-(1-\sqrt{2})^x=\frac{3}{2}$
How do I found out the final solution? Thank you very much!
P.S. The answers I can choose from are:
a) $x=1$
b) $x=2$
c) $x=\frac{2\lg2}{\lg(3+2\sqrt2)}$
d) $x=\frac{2\lg2}{\lg(3-2\sqrt2)}$
e) no solution
f) $x=2\lg2$
 A: First of all, $\sqrt{3-2\sqrt{2}}$ is $\sqrt{2} - 1$.
Now, Let $(\sqrt{2}+1)^x$ be y.
$y - \frac{1}{y} = \frac{3}{2}$.
So $2y^2 - 3y - 2 = 0$.
So y is 2 or $-\frac{1}{2}$. As y = $(\sqrt{2}+1)^x$, y must be positive and thus 2.
$(\sqrt{2}+1)^x = 2$
So $x = \frac{log(2)}{log(\sqrt{2}+1)}$, which is same as option c) here as you can see by simplifying the expression in c). I guess they are also taking lg instead of log, but as you can see it makes no difference whatsoever.
A: $$(\sqrt{3+2\sqrt{2}})^x-(\sqrt{3-2\sqrt{2}})^x=\frac{3}{2}$$
Multiply both side with $(\sqrt{3+2\sqrt{2}})^x$, then you get
$$(\sqrt{3+2\sqrt{2}})^{2x}-(\sqrt{(3-2\sqrt{2})(3+2\sqrt{2})})^x=\frac{3}{2}(\sqrt{3+2\sqrt{2}})^x$$
$$(\sqrt{3+2\sqrt{2}})^{2x}-(1)^x=\frac{3}{2}(\sqrt{3+2\sqrt{2}})^x$$
Let  $y =(\sqrt{3+2\sqrt{2}})^x$, then 
  $y^2 -1 = \frac{3}{2}y$ 
Solve above equation, $y =2$ or $y = \frac{-1}{2}$(its never true)
Therefore, $y =2$ that implies $(\sqrt{3+2\sqrt{2}})^x =2$
Apply $log$ both side,$$x\log(\sqrt{3+2\sqrt{2}}) =log(2)$$
Therefore, $$x=\frac {2\log(2)}{\log({3+2\sqrt{2}})}$$
