1. ## solving exponential equation

$\displaystyle 4^x*\frac{1}{2}^{3-2x}=8*(2^x)^2$

This is as far as i got:
$\displaystyle \frac{1}{2}^{-2x}*\frac{1}{2}^{3-2x}=\frac{1}{2}^{-3}*\frac{1}{2}^{-x^2}$

$\displaystyle 2x*3-2x=3*-x^2$

$\displaystyle 6x-2x=-3x^2$

$\displaystyle 3x^2+4x=0$
any help would be much appriciated!

2. Factor the x out front, and then solve each of the two factors.

3. Originally Posted by windir
$\displaystyle 4^x*\frac{1}{2}^{3-2x}=8*(2^x)^2$

This is as far as i got:
$\displaystyle \frac{1}{2}^{-2x}*\frac{1}{2}^{3-2x}=\frac{1}{2}^{-3}*\frac{1}{2}^{-x^2}$

$\displaystyle 2x*3-2x=3*-x^2$

$\displaystyle 6x-2x=-3x^2$

$\displaystyle 3x^2+4x=0$
any help would be much appriciated!
$\displaystyle 4^x \cdot \left(\frac{1}{2}\right)^{3-2x} = 8 \cdot 2^{2x}$

$\displaystyle 2^{2x} \cdot 2^{2x-3} = 8 \cdot 2^{2x}$

$\displaystyle 2^{2x-3} = 8$

$\displaystyle 2^{2x-3} = 2^3$

$\displaystyle 2x-3 = 3$

$\displaystyle x = 3$

4. oh wow. thank you so much! i cant believe i didnt see that before!