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- Sep 6th 2007, 05:38 PMRevelsynExponent problem!!
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- Sep 6th 2007, 06:05 PMtopsquark
$\displaystyle 2^{-n} \left ( 2^n - 2^{1 + n} \right )$

First, note that multiplication is distributive over addition, so this is:

$\displaystyle 2^{-n} \left ( 2^n - 2^{1 + n} \right ) = 2^{-n} \cdot 2^n - 2^{-n} \cdot 2^{1 + n}$

Now use the property that

$\displaystyle a^x \cdot a^y = a^{x + y}$

So:

$\displaystyle = 2^{-n + n} - 2^{-n + 1 + n} = 2^0 - 2^1$

$\displaystyle = 1 - 2 = -1$

-Dan - Sep 6th 2007, 06:08 PMKrizalid
Other way

$\displaystyle 2^{-n}(2^n-2^{1+n})=2^{-n}\cdot2^n(1-2)=-1$

Cheers,

K.