[tex]a)\,\dfrac{1}{16}=16^{-1}=(2^4)^{-1}=2^{-4}\,\,\checkmark\\\\ b) 1024=2^{10}\,\,\checkmark\\\\ c) 5\sqrt{8}=5.\sqrt{2^3}=5.2^{\frac32}\,\,\checkmark\\\\\ d) 16\sqrt{32}=2^{4}.\sqrt{2^5}=2^{4}.2^{\frac52}=2^{\frac{13}{2}}\,\,\checkmark\\\\ e) \left(\sqrt{\dfrac{128}{2}}\right)^{-3}=(\sqrt{64})^{-3}=8^{-3}=(2^3)^{-3}=2^{-9}\,\,\checkmark[/tex]
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Resposta:
c) 128/2-³ = 2) Coloque em forma de potencia de base 3: a) ³√9 = b) 3⁵√27/243
Explicação passo a passo:
[tex]a)\,\dfrac{1}{16}=16^{-1}=(2^4)^{-1}=2^{-4}\,\,\checkmark\\\\ b) 1024=2^{10}\,\,\checkmark\\\\ c) 5\sqrt{8}=5.\sqrt{2^3}=5.2^{\frac32}\,\,\checkmark\\\\\ d) 16\sqrt{32}=2^{4}.\sqrt{2^5}=2^{4}.2^{\frac52}=2^{\frac{13}{2}}\,\,\checkmark\\\\ e) \left(\sqrt{\dfrac{128}{2}}\right)^{-3}=(\sqrt{64})^{-3}=8^{-3}=(2^3)^{-3}=2^{-9}\,\,\checkmark[/tex]
É isso!!