Exponential equation help

1. Jul 17, 2008

babacanoosh

Hello all, I can't believe I having a hard time with this problem but...I am

1. 2^(x^2) X 4^(2x) = 1/8

2. make the bases the same?

3.
2^(x^2) X 2^((2x)^2) = 1/8
2^(3x^(2)) = 1/8
3x^(2)log2=log1/8
(log1/8)/(log2) = 3x^(2)
x^(2) = -1

I know I am doing it wrong. Any help would be greatly appreciated. Thanks,
Baba

2. Jul 17, 2008

rock.freak667

$$(2^{2x})^2=2^{4x}$$

3. Jul 17, 2008

dynamicsolo

The general rule for "exponentiated exponents" is

$$(a^b)^c = a^{(b \cdot c)$$, for a positive base a .

For this problem, you will end up with a quadratic equation in x, with two solutions.

Last edited: Jul 17, 2008
4. Jul 17, 2008

babacanoosh

great! thanks so much. Just a stupid mistake on my part, I knew that :uhh:

5. Jul 18, 2008

epkid08

I believe you end up with two imaginary solutions; My guess is that you should just show your work until you run into negative square root.

6. Jul 18, 2008

rock.freak667

You actually get two real roots.

7. Jul 18, 2008

lukas86

I found the same, you should get two real roots.

8. Jul 18, 2008

epkid08

Oh, I see what I did wrong, I had $$2^{x^2}(2^x + 2^x)=1/8$$, instead of $$2^{x^2}*2^x*2^x=1/8$$