How would I solve this?

2^2x - 2^(x+1) = 0

Having trouble with this too

2^2x - 12 * 2^x + 32 = 0

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- Sep 6th 2006, 09:27 AMclassicstringsLogarithm
How would I solve this?

2^2x - 2^(x+1) = 0

Having trouble with this too

2^2x - 12 * 2^x + 32 = 0 - Sep 6th 2006, 09:38 AMGlaysher
$\displaystyle 2^{2x}=2^{x+1}$

$\displaystyle 2x=x+1$

$\displaystyle x=1$ - Sep 6th 2006, 09:41 AMGlaysher
For second one

$\displaystyle (2^x - 8)(2^x - 4) = 0$

$\displaystyle 2^x =8$ gives $\displaystyle x=3$

$\displaystyle 2^x =4$ gives $\displaystyle x=2$ - Sep 7th 2006, 08:27 AMclassicstrings
Thanks Glaysher!