1. ## solve 3

2. I'm not completely sure how to do this as I han't covered it :P
but here's some aid:

from that we get: (I'm not entirely sure)

1^x + 1^x + 6^x + 12^x = 32 (by adding/subtracting in the fractions)
and 1^anything is always 1
so

1 + 1 + 6^x + 12^x = 32
2 + 6^x + 12^x = 32
6^x + 12^x = 30

and that's all I've gotten up to since I haven't done this before

hope someoone else can help and please correct me if I've done anything wrong

3. I think you might be mistaken in that

$6^{\frac{1}{2}}=\sqrt{6} \not= 6 + \frac{1}{2}$

The whole $\frac{1}{2}$ is an exponent, and they can't be added or subtracted easily.
By the way, looks like this is a really hard problem :/

4. Originally Posted by DivideBy0
I think you might be mistaken in that

$6^{\frac{1}{2}}=\sqrt{6} \not= 6 + \frac{1}{2}$

The whole $\frac{1}{2}$ is an exponent, and they can't be added or subtracted easily.
By the way, looks like this is a really hard problem :/
oh sorry, I thought it meant 1/2
thanks for telling me

5. Hello, perash!

I have an "eyeball" solution . . .

$\left(6^{\frac{1}{2}} - 5^{\frac{1}{2}}\right)^x + \left(3^{\frac{1}{2}}
- 2^{\frac{1}{2}}\right)^x + \left(3^{\frac{1}{2}} + 2^{\frac{1}{2}}\right)^x + \left(6^{\frac{1}{2}} + 5^{\frac{1}{2}}\right)^x \;=\;32$
We have: . $\left(\sqrt{6}-\sqrt{5}\right)^x + \left(\sqrt{6} + \sqrt{5}\right)^x + \left(\sqrt{3} - \sqrt{2}\right)^x + \left(\sqrt{3} + \sqrt{2}\right)^x \;=\;32$

Since the left side equals an integer, $x$ must be even: . $x = 2n$

We have: . $\left(\sqrt{6}-\sqrt{5}\right)^{2n} + \left(\sqrt{6} + \sqrt{5}\right)^{2n} + \left(\sqrt{3} - \sqrt{2}\right)^{2n} + \left(\sqrt{3} + \sqrt{2}\right)^{2n} \;=\;32$

. . $\left[\left(\sqrt{6}-\sqrt{5}\right)^2\right]^n + \left[\left(\sqrt{6} + \sqrt{5}\right)^2\right]^n + \left[\left(\sqrt{3} - \sqrt{2}\right)^2\right]^n + \left[\left(\sqrt{3} + \sqrt{2}\right)^2\right]^n \;=\;32$

. . $\left(11 - 2\sqrt{30}\right)^n + \left(11 + 2\sqrt{30}\right)^n + \left(5 - 2\sqrt{6}\right)^n + \left(5 + 2\sqrt{6}\right)^n \;=\;32$

And we see that $n = 1$ is a solution.

Therefore: . $x \:=\:2n \:=\:\boxed{2}$

6. Graphing it shows that -2 is also a value

7. Originally Posted by perash
Graphing it shows that -2 is also a value
In retrospect we should have expected that because:
$\sqrt{6} - \sqrt{5} = \frac{\sqrt{6} - \sqrt{5}}{1} \cdot \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}}$

$= \frac{6 - 5}{\sqrt{6} + \sqrt{5}} = \frac{1}{\sqrt{6} + \sqrt{5}}$

The same holds for the $\sqrt{3} - \sqrt{2}$ term so we have:
$(\sqrt{6} - \sqrt{5})^x + (\sqrt{6} + \sqrt{5})^x + (\sqrt{3} - \sqrt{2})^x + (\sqrt{3} + \sqrt{2})^x$

$= \frac{1}{(\sqrt{6} + \sqrt{5})^x} + (\sqrt{6} + \sqrt{5})^x + \frac{1}{(\sqrt{3} + \sqrt{2})^x} + (\sqrt{3} + \sqrt{2})^x$

So if we have a + x solution, we should also have the - x solution also.

-Dan