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Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0Why don't we look at it, from the very proof, let's say we have any two numbers with different exponents: \[a^m . a^n\] By definition, they represent a sucession of products so therefore: \[\left[ a.a.a.a.a.a.... (m.factors) \right]\left[ a.a.a.a.a...(n.factors) \right]\] a in multiplied "m" times, and also "n" times, so we can deduce that a is multiplied "m+n" times: \[\left[ a.a.a.a.a.a....(m+n.factors) \right]\] Therefore, by the very definition of exponent: \[a ^{m+n}\] Giving us the conclusion: \[a^m. a^n = a ^{m+n}\]

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0This means that if you have: \[4^2 . 4^8 = 4^{2+8}\] I presume you know what 2+8 is.
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