Good day everyone,
Does anyone know how to solve this,
Given $\displaystyle 2^{2x}+64(2^{-x})= 32$. Find the value of x.
Thanks in advance!
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
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 $\displaystyle u^2 + 4u - 16 = 0$