Question

calcue a soma de S : log³√9 √1/27 - log³√0,5 √8 + log³√100 raiz de 6√0,1

Answer 1

Resposta:

2

Explicação passo-a-passo:

Oi! Para resolver esse exercício, iremos calcular o valor de cada parcela dessa expressão individualmente e depois somaremos os resultados.

Mas antes, temos que conhecer as seguintes informações:

1. $\log_ab=c~\leftrightarrow~a^c=b$

2. $\log_ab^k=k\log_ab$

3. $\sqrt[a]{x^b}=x^{\frac{b}{a}}$

4. $(x^a)^b=x^{ab}$

Agora, vamos começar!

$\log_{\sqrt[3]{9}}{\sqrt{\frac{1}{27}}}}=x\\\\\log_{9^{\frac{1}{3}}}{(\frac{1}{27})^{\frac{1}{2}}}=x\\\\\log_{(3^2)^{\frac{1}{3}}}{(3^{-3})^{\frac{1}{2}}}=x\\\\\log_{3^{\frac{2}{3}}}{3^{-\frac{3}{2}}}=x\\\\(3^{\frac{2}{3}})^x=3^{-\frac{3}{2}}\\\\3^{\frac{2x}{3}}=3^{-\frac{3}{2}}\\\\\frac{2x}{3}=-\frac{3}{2}\\\\2x=-\frac{9}{2}\\\\\boxed{x=-\frac{9}{4}}$

$-\log_{\sqrt[3]{0,5}}{\sqrt{8}}=y\\\\-\log_{0,5^{\frac{1}{3}}}{8^{\frac{1}{2}}}=y\\\\-\log_{(\frac{1}{2})^{\frac{1}{3}}}{(2^3)^{\frac{1}{2}}}=y\\\\-\log_{(2^{-1})^{\frac{1}{3}}}{2^{\frac{3}{2}}}=y\\\\\log_{2^{-\frac{1}{3}}}{(2^{\frac{3}{2}})^{-1}}=y\\\\\log_{2^{-\frac{1}{3}}}{2^{-\frac{3}{2}}}=y\\\\(2^{-\frac{1}{3}})^y=2^{-\frac{3}{2}}\\\\2^{-\frac{y}{3}}=2^{-\frac{3}{2}}\\\\-\frac{y}{3}=-\frac{3}{2}\\\\\frac{y}{3}=\frac{3}{2}\\\\\boxed{y=\frac{9}{2}}$

$\log_{\sqrt[3]{100} }{\sqrt[6]{0,1} }=z\\\\\log_{100^{\frac{1}{3}}}{0,1^{\frac{1}{6}}}=z\\\\\log_{(10^2)^{\frac{1}{3}}}{(10^{-1})^{\frac{1}{6}}}=z\\\\\log_{10^{\frac{2}{3}}}{10^{-\frac{1}{6}}}=z\\\\(10^{\frac{2}{3}})^z=10^{-\frac{1}{6}}\\\\10^{\frac{2z}{3}}=10^{-\frac{1}{6}}\\\\\frac{2z}{3}=-\frac{1}{6}\\\\2z=-\frac{3}{6}\\\\2z=-\frac{1}{2}\\\\\boxed{z=-\frac{1}{4}}$

Efetuando x + y + z, obtemos:

$-\dfrac{9}{4}+\dfrac{9}{2}-\dfrac{1}{4}=\dfrac{-9+18-1}{4}=\dfrac{8}{4}=\boxed{2}$

Saiba mais em:

Propriedades de log: brainly.com.br/tarefa/25620028

Espero ter ajudado. Qualquer dúvida pode deixar nos comentários. Bons estudos! ♥️