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#1

$Old$ 11-21-2008, 01:12 AM

Kitty216 $Kitty216 is offline$

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$Default$ solving for x

Solve for x

It looks so simple but when I tried it, I ended up getting strange numbers. Help!

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#2

$Old$ 11-21-2008, 02:31 AM

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Solve for x

It looks so simple but when I tried it, I ended up getting strange numbers. Help!

Here's an outline - you can fill in the details.

You have $(2^2)^{2x} - 2 \cdot (2^2)^{x+4} + (2^2)^8 = 0$ .

Substitute $w = 2^x$ and do a bit of simplifying.

You end up with $w^2 - 2^9 w + 2^{16} = 0$ .

This factorises: $(w - 2^8)^2 = 0$ .

Therefore $2^x = 2^8$ .

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Last edited by mr fantastic; 11-21-2008 at 02:45 AM.

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$Old$ 11-21-2008, 02:42 AM

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I'm kinda confused how you substituted

in and got

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$Old$ 11-21-2008, 02:48 AM

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Originally Posted by Kitty216 $View Post$

I'm kinda confused how you substituted

in and got

$(2^2)^{2x} - 2 \cdot (2^2)^{x+4} + (2^2)^8 = 0$

From the usual index laws:

$\Rightarrow (2^{2x})^2 - 2^{2x + 9} + 2^16 = 0$

$\Rightarrow (2^{2x})^2 - 2^9 \cdot 2^{2x} + 2^16 = 0$

Now make the substitution.

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#5

$Old$ 11-21-2008, 03:01 AM

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ohh okay, now I got

how do I factor that?

doesn't seem to work.

ahhh.. nvm! I got it now! It does work! =D Thanks so much!!

Last edited by Kitty216; 11-21-2008 at 03:06 AM. Reason: got the answer

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