How do I solve an exponential equation with different bases?

[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

 

[rock.freak667]

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

 

[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.

 

[babacanoosh]

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

 

[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.

 

[rock.freak667]

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.

You actually get two real roots.

 

[lukas86]

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

 

[epkid08]

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