(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
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)
$\displaystyle (2^{100}) - (2^{99})$
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:
$\displaystyle 2^{100} - 2^{99}$
$\displaystyle 2(2^{99})-2^{99}$
$\displaystyle =1(2^{99})$
$\displaystyle =2^{99}$