Equation

PLEASE, /sci/, I need your help!

Here's the equation:
8*3^{x+\sqrt{x}}=9^x-9^{\sqrt{x}+1}

I can't solve it. I tried many things, but I fail every single time. I know the answer (which Mathematica/Wolfram|Alpha gave me) but that's not what I want.
What I need is a step-by-step solution or some good hints.

I'll be very glad if you help me out.
Thanks in advance.

Damn, forgot about the brackets.

8*3^{x+\sqrt{x}}=9^x-9^{\sqrt{x}+1}

Write \alpha = 3^x and \beta = 3^{\sqrt{x}}.  Then the equation is

8\alpha \beta = \alpha^2 -9\beta^2
0= \alpha^2 -8\alpha \beta-9\beta^2
0= (\alpha - 9\beta)(\alpha +\beta)

So 3^x = 9\cdot 3^\sqrt{x} or 3^x = - 3^\sqrt{x}.  These should be easy to solve with logarithms and such.

3^x = -3^sqrt{x}
Well this equation has no solutions, hasn't it?

I meant 3^x = -3^/sqrt{x} ...

I can't get used to this TeX :P

3^x = -3^\sqrt{x}

Protip: use mathbin.net to preview your TeX shit.

Protip: Spambomb moot until he gets his lazy ass together and puts in a Delete feature on the textboards. :/

>>6
Sure it does.

\log 3^x = \log -3^{\sqrt{x}}
x = \log {-1} + \sqrt{x}  + 2\pi i n
\sqrt{x}^2 - \sqrt{x} - (2n+1) \pi i = 0
\sqrt{x} = \frac{1 \pm \sqrt{1+4(2n+1)\pi i}}{2}
x = \left[\frac{1 \pm \sqrt{1+4(2n+1)\pi i}}{2}\right]^2

>>8
Hmm.. That's something I don't understand yet. How did you get the second line?
And is it actually possible to logarithm a negative number?

>>9
The "-" signs in the first 2 lines are negative signs, not subtraction.

And you can do logs of negative numbers, but you have to use complex numbers.  Also, there's more than one logarithm for any number.  So the logarithms of 1 are 2\pi i n for any integer n, since e^{2\pi in} = 1.

http://en.wikipedia.org/wiki/Complex_logarithm

>>9
http://en.wikipedia.org/wiki/Complex_logarithm