(2^100) - (2^99)

im supposed to be doing pretty advance stuff but this just stumps me

can someone show me how this is done? the answer is to be in the form 2^n

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- Feb 10th 2010, 01:08 AMfuror celticaindice thing 1
(2^100) - (2^99)

im supposed to be doing pretty advance stuff but this just stumps me

can someone show me how this is done? the answer is to be in the form 2^n - Feb 10th 2010, 01:47 AMAwsom Guy
ok not too hard. the answer is 2^1. When your adding indices you add the top and when your subtracting them you subtract the indices. Try check this website out:

Subtracting Indices (with worked solutions & videos) - Feb 10th 2010, 03:11 AMfuror celtica
thanks for trying to help kid but you are really wrong.(Crying) i think you should review your lessons and check out the reliability of your websites, i think anybody could see this right away.

- Feb 10th 2010, 07:21 AMQuacky
$\displaystyle (2^{100}) - (2^{99})$

I'd take out a common factor of $\displaystyle 2^{99}$

$\displaystyle =2^{99}(2-1)$

$\displaystyle =2^{99}(1)$

This will need verification.

Actually, I am sure that I am correct.

$\displaystyle 2\times2^{99}=2^{100}$

Substituting this into the first equation:

$\displaystyle 2^{100} - 2^{99}$

$\displaystyle 2(2^{99})-2^{99}$

$\displaystyle =1(2^{99})$

$\displaystyle =2^{99}$ - Feb 10th 2010, 07:25 AMQuacky
I had to make quite a few edits due to my rushed answer. The correct answer is displayed above.