Hi. I'm having trouble trying to solve the following equation for $\displaystyle x$
$\displaystyle 2^{2x}-12\times2^x+32=0$
Any help will be appreciated. THanks
Let $\displaystyle X = 2^x$, so that the equation becomes
$\displaystyle X^2 - 12X + 32 = 0$
$\displaystyle (X - 8)(X - 4) = 0$
$\displaystyle X = 8$ or $\displaystyle X = 4$.
Thus $\displaystyle 2^x = 8$ or $\displaystyle 2^x = 4$.
It should be pretty obvious that $\displaystyle x = 3$ or $\displaystyle x = 2$.
for this question, you don't need to use logarithm.
$\displaystyle let 2^x be y.$
$\displaystyle y^2-12y+32=0$
$\displaystyle (y-4)(y-8)=0$
y=4 or y=8
$\displaystyle 2^x=4$ or $\displaystyle 2^x=8$
$\displaystyle 2^x=2^2$ or $\displaystyle 2^x=2^3$
By comparing indices,
$\displaystyle x=2$ or $\displaystyle 3$