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November 21st, 2008, 02:48 AM

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Quote:

Originally Posted by Kitty216

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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