# solving exponential equation

• Oct 7th 2009, 02:36 PM
windir
solving exponential equation
$4^x*\frac{1}{2}^{3-2x}=8*(2^x)^2$

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

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

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

$3x^2+4x=0$
any help would be much appriciated!
• Oct 7th 2009, 02:45 PM
stapel
Factor the x out front, and then solve each of the two factors. (Wink)
• Oct 7th 2009, 02:45 PM
skeeter
Quote:

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

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

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

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

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

$4^x \cdot \left(\frac{1}{2}\right)^{3-2x} = 8 \cdot 2^{2x}$

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

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

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

$2x-3 = 3$

$x = 3$
• Oct 7th 2009, 02:51 PM
windir
oh wow. thank you so much! i cant believe i didnt see that before!