1. ## Logarithmic Equation

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

2. Originally Posted by Stroodle
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 $\displaystyle x = 2$.

3. Originally Posted by Stroodle
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
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 $\displaystyle 3$

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...