Question

s= log na base 10 de 0,001 + log na base 3 de 3 raiz quadrada de 3 - log na base 8 de 16

Answer 1

Resolvendo cada logaritmo separadamente:

$\log_{10}0,001=\dfrac{\log 0,001}{\log 10}\\\\ \log_{10}0,001=-3\\\\\\\\ \log_{3}3\sqrt{3}=\log_33+\log_3\sqrt{3}\\\\ \log_33=1\\ \log_3\sqrt{3} \to 3^x=3^{\frac{1}{2}} \to x=\frac{1}{2}\\\\ \log_{3}3\sqrt{3}=1+\frac{1}{2}=\frac{3}{2}$



$\log_{8}16\to 8^x=16\\\\ (2^3)^x=2^4\\\\ 2^{3x}=2^4\\\\ 3x=4\\\\ x=\frac{4}{3}\\\\ \log_816=\frac{4}{3}$



Calculando tudo a partir dos resultados encontrados:

$S=-3+\frac{3}{2}-\frac{4}{3}\\\\ S=\frac{-18+9-8}{6}\\\\ \boxed{S=-\frac{17}{6}}$


Bons estudos!