Question

Resolva essas questões de função logaritma:

Answer 1

Resposta:

Explicação passo-a-passo:

1)

$\frac{log _{3 ^{1} + log _10 ^{0,01} } }{log _2 ^{\frac{1}{64} . log _4 ^{\sqrt{8} }}} \\\\ \frac{{0 + log _{10} ^{10^{-2} } } }{log _2 ^{2^{-6} } . log _4 ^{2^{\frac{3}{2} } } }}}\\\\\frac{- 2}{- 6 . \frac{3}{4} } =\\\\\frac{- 2}{\frac{-18}{4} } =\\\\\ - 2 . {\frac{-4}{18} } =\\\\\frac{ 8}{18} =\\\frac{4}{9}$

2)

$log_{3} ^{x} + log _{9} ^{x} = 1\\\\\\log_{3} ^{x} + \frac{log_{3} ^{x} }{log_{3} ^{9} } = 1\\\\log_{3} ^{x} + \frac{log_{3} ^{x} }{2 } = 1\\\\2. log_{3} ^{x} + log_{3} ^{x} = 2\\3 . log_{3} ^{x} = 2\\log_{3} ^{x} = \frac{2}{3} \\\\x = 3^{\frac{2}{3}} \\x = \sqrt[3]{3^{2} } \\x = \sqrt[3]{9}$