Hey! I'm back again!Itseems so simple!(Worried)

How can I do this:::

http://img516.imageshack.us/img516/8240/mathxf1.png

http://img516.imageshack.us/img516/m...png/1/w451.png

See you!!!

Printable View

- Sep 30th 2008, 05:02 AMWithzAlgebra!!!!!!!!!!
Hey! I'm back again!

**It****seems so simple!**(Worried)

How can I do this:::

http://img516.imageshack.us/img516/8240/mathxf1.png

http://img516.imageshack.us/img516/m...png/1/w451.png

See you!!! - Sep 30th 2008, 05:57 AMChop Suey
Note that $\displaystyle 4^{x} + 4^{-x} = 2^{2x} + 2^{-2x}$

And we know that $\displaystyle 2^{x} + 2^{-x} = 3$. By squaring both sides we get:

$\displaystyle 2^{2x} + 2(2^{0}) + 2^{-2x} = 9$ - Sep 30th 2008, 11:05 AMWithzthanx
Yeah! That's it!

Thank you so much!!! (Clapping) It Helps me a lot!!!