Question 88735
{{{2^8/2^3}}}


When you divide common bases with exponents, you subtract the exponents


{{{2^8/2^3=2^(8-3)=2^5=32}}}


Notice that {{{2^8/2^3}}} really looks like


{{{2^8/2^3=(2*2*2*2*2*2*2*2)/(2*2*2)}}}


{{{2^8/2^3=(cross(2*2*2)*2*2*2*2*2)/cross(2*2*2)}}} When you divide, it's like you're taking away 3 "2"s from the numerator. So that is the reason why you subtract the exponents.

{{{2^8/2^3=(cross(2*2*2)*2*2*2*2*2)/cross(2*2*2)=2^5}}} So you're left with 5 "2"s, or {{{2^5}}}