zeltzinmg zeltzinmg
  • 04-09-2018
  • Mathematics
contestada

explain your reasoning as you compare the values of a^n and a^-n when n<0 brainly

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jdoe0001 jdoe0001
  • 04-09-2018

n < 0, is another way to say "n is negative", so let's check


[tex] \bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a^n~\hspace{10.5em}\stackrel{n = -n}{a^{-n}}\implies \cfrac{1}{a^n} \\\\\\ a^{-n}~\hspace{10em}\stackrel{n=-n}{a^{-(-n)}}\implies a^{+n}\implies a^n [/tex]

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