Question

ALGUÉM ME AJUDA PFVV!! É pra hoje!

Answer 1

Sabendo que $\log_ab=y$ é equivalente à $a^y=b$

$\log_20,5=y\\\\2^y=0,5\\\\2^y=\frac{1}{2}\\\\ 2^y=2^-^1\\\\y=-1$

Utilizei a propriedade $\frac{1}{a^b}=a^-^b$

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$\log_3\sqrt3=y'\\\\3^y'=\sqrt3\\\\3^y'=3^\frac{1}2}\\\\y'=\frac{1}2}$

Utilizei a propriedade $a^\frac{b}{c}=\sqrt[c]{a^b}$

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$\log_48=y''\\\\4^y''=8\\\\(2^2)^y''=2^3\\\\2^2^y''=2^3\\\\2y''=3\\\\y''=\frac{3}{2}$

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$y+y'+y''\\\\-1+\frac{1}2}+\frac{3}{2}\\\\\frac{-2+1+3}{2}=\frac{2}{2} = 1$