Question

log
$\sqrt[3]{25}$
na base 1/5+log0,01​

Answer 1

Explicação passo-a-passo:

Vamos calcular os log separados

$\log_{1\over5}\sqrt[3]{25} =x\\ \\ ({1\over5})^x=\sqrt[3]{5^2} \\ \\ 5^{-x}=5^{2\over3}\\ \\ cancela~~base~~5\\ \\ -x={2\over3}\\ \\ x=-{2\over3}$

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$log0,01=x\\ \\ 10^x=0,01\\ \\ 10^x=10^{-2}\\ \\ cancela~~ a ~~base~~10\\ \\ x=-2$

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$Calculando\\ \\ -{2\over3}-2={-2-6\over3}=\fbox{ -{8\over3} }$