# Thread: problem solving for algebra

1. ## problem solving for algebra [solved]

Good day everyone,

Does anyone know how to solve this,

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

2. Originally Posted by imforumer
Given $\displaystyle 2^{2x}+64(2^{-x})= 32$. Find the value of x.
Welcome to the forum Imforumer

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

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

let $\displaystyle 2^x = u$

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

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

Can you finish ?

Bobak

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

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

4. Originally Posted by imforumer
I also can't solve it, I stuck at where you stuck.

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

If 4 is a root, then $\displaystyle u-4$ is a factor of $\displaystyle u^3 - 32u + 64$.So do polynomial devision and find the quotient.
$\displaystyle u^3 - 32u + 64 = (u-4)(u^2 + 4u - 16) = 0$
Surely you should know how to solve the quadratic $\displaystyle u^2 + 4u - 16 = 0$