problem solving for algebra

imforumer · #1 $Old$ June 22nd, 2008, 03:07 AM

Good day everyone,

Does anyone know how to solve this,

Given $2^{2x}+64(2^{-x})= 32$ . Find the value of x.

Thanks in advance!

bobak · #2 $Old$ June 22nd, 2008, 03:14 AM

Quote:

Originally Posted by imforumer $View Post$

Given $2^{2x}+64(2^{-x})= 32$ . Find the value of x.

Welcome to the forum Imforumer

This first question multiply through the equation by $2^x$

$2^{3x}- 32 \cdot 2^{x}+64=0$

let $2^x = u$

$u^3 - 32 u+64=0$

Notice that $u = 4$ is a solution to this equation.

Can you finish ?

Bobak

imforumer · #3 $Old$ June 22nd, 2008, 05:52 AM

I also can't solve it, I stuck at where you stuck.

Btw, the answer given is $x=1.31$ (3 significant figure), $x=2$ , no solution is provided =.="

thanks bobak for your time.

Isomorphism · #4 $Old$ June 22nd, 2008, 06:22 AM

Quote:

Originally Posted by imforumer $View Post$

I also can't solve it, I stuck at where you stuck.

Btw, the answer given is $x=1.31$ (3 significant figure), $x=2$ , no solution is provided =.="

thanks bobak for your time.

If 4 is a root, then $u-4$ is a factor of $u^3 - 32u + 64$ .So do polynomial devision and find the quotient.

$u^3 - 32u + 64 = (u-4)(u^2 + 4u - 16) = 0$

Surely you should know how to solve the quadratic $u^2 + 4u - 16 = 0$