1. ## Logarithmic Equation

Hi. I'm having trouble trying to solve the following equation for $x$

$2^{2x}-12\times2^x+32=0$

Any help will be appreciated. THanks

2. Originally Posted by Stroodle
Hi. I'm having trouble trying to solve the following equation for $x$

$2^{2x}-12\times2^x+32=0$

Any help will be appreciated. THanks
Let $X = 2^x$, so that the equation becomes

$X^2 - 12X + 32 = 0$

$(X - 8)(X - 4) = 0$

$X = 8$ or $X = 4$.

Thus $2^x = 8$ or $2^x = 4$.

It should be pretty obvious that $x = 3$ or $x = 2$.

3. Originally Posted by Stroodle
Hi. I'm having trouble trying to solve the following equation for $x$

$2^{2x}-12\times2^x+32=0$

Any help will be appreciated. THanks
for this question, you don't need to use logarithm.

$let 2^x be y.$
$y^2-12y+32=0$
$(y-4)(y-8)=0$

y=4 or y=8
$2^x=4$ or $2^x=8$
$2^x=2^2$ or $2^x=2^3$
By comparing indices,
$x=2$ or $3$

4. Thanks heaps for your help guys. Knew it would be something simple that I missed.
Was in the logarithm section, so I was trying to do it with logs...