Question

Determine o valor de E = Log 3√64 na base 2 - log 1 na base 8 + log de 27/64 na base 4/3

Answer 1

$E=log_2(\sqrt[3]{64})-log_81+log_{\frac{4}{3}}(\frac{27}{64})\\ E=log_2(\sqrt[3]{4^3})-0+log_{\frac{4}{3}}(\frac{3^3}{4^3})\\ E=log_24+log_{(\frac{3}{4})^{-1}}(\frac{3}{4})^3\\\\ Calculando\ log_24:\\\\ 2^x=4\ --\ \textgreater \ \ 2^x=2^2\\ x=2 \\\\ Calculando\ log_{(\frac{3}{4})^{-1}}(\frac{3}{4})^3:\\\\ ((\frac{3}{4})^{-1})^x=(\frac{3}{4})^3\ --\ \textgreater \ \ (\frac{3}{4})^{-x}=(\frac{3}{4})^3 \\ -x=3\ --\ \textgreater \ \ x=-3 \\\\ Logo:\\\\ E=2+(-3)\\ E=2-3\\ \| \ E=-1\ \|$

Answer 2

gente como assim, vo me mata