Question

Usando as propriedades,calcule:

log₂ 16√8 =

log₃ 81√3 =
         ∛3
    
 Alguém me ajude,por favor!

Answer 1

Lembrando:

$\sqrt[n]{a^{x}}=a^{x/n}$

$log_{n}~n= 1$
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$log_{2}(16\sqrt{8})=log_{2}(16\sqrt[2]{2^{3}}) \\ log_{2}(16\sqrt{8})=log_{2}(2^{4}*2^{3/2}) \\ log_{2}(16\sqrt{8})=log_{2}(2^{4+[3/2]})\\log_{2}(16\sqrt{8})=log_{2}(2^{11/2})\\log_{2}(16\sqrt{8}=(11/2)*log_{2}(2)\\log_{2}(16\sqrt{8})=(11/2)*1\\log_{2}(16\sqrt{8})=11/2$
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$log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{4}*3^{1/2}/3^{1/3})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{4+[1/2]-[1/3]})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{[24+3-2]/6})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{25/6})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=(25/6)*log_{3}(3)\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=(25/6)*1\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=25/6$