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.


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) $x=2\lg2$

  • $\begingroup$ The key structural thing is that $3-2\sqrt{2}=\frac{1}{3+2\sqrt{2}}$. $\endgroup$ – André Nicolas Jul 18 '12 at 12:35

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.

  • $\begingroup$ Can you please explain why the $(\sqrt2-1)^x$ is $\frac{1}{y}$ and the transformation of x into $log$? Thank you very much for your help! $\endgroup$ – Grozav Alex Ioan Jul 18 '12 at 9:52
  • $\begingroup$ $\sqrt{2}-1$ is inverse of $\sqrt{2}+1$, so $(\sqrt{2}-1)^x$ is inverse of $(\sqrt{2}+1)^x$. For the other question, if $a^x=b$, then by taking log on both sides, $x log(a) = log(b)$. Therefore $x = \frac{log(b)}{log(a)}$. $\endgroup$ – Wonder Jul 18 '12 at 9:57


Multiply both side with $(\sqrt{3+2\sqrt{2}})^x$, then you get



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, $$x=\frac {2\log(2)}{\log({3+2\sqrt{2}})}$$


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