2^{2x+1}- 5 * 2^{x}- 12 = 0

If 2^{x}is y, then it would be like this:

y^{2}- 5y - 12 = 0

Seems simple enough, but:

$\displaystyle \frac{-5 \pm \sqrt{5^{2} - 4 * 1 * -12 }}{2} $

equals:

$\displaystyle \frac{-5 \pm \sqrt{73 }}{2} $

And there is no string square root of 73. Is there a way to get around this, some trick perhaps?

Next one is:

$\displaystyle 5^{x^{2} - 3x + 1} = \frac{1}{5} $

I believe I have to solve the $\displaystyle x^{2} - 3x + 1 $ part first, and so I tried, but encountered the same problem as above: The square root ends up like a decimal - which shouldn't happen :/

And neither last nor least, is:

$\displaystyle (\frac{1}{2})^{x} * (\frac{2}{3})^{x+1} * (\frac{3}{4})^{x+2} = 6 $

I don't even know how to start this one. I believe that's about it from exponentials, but I have to cover two more topics about trigonometry :O

Thanks