Question

Use a definição de logaritmo para calcular o valor de E = log 125/345 1,4 - log100000 0,1 - log 1/5

ajuda pf

Answer 1

Resposta:

a) $-\frac{1}{10}$

b) $\sqrt{2}$

Explicação passo a passo:

a)

→ $\log_{\frac{125}{343}}{1,4}=\log_{\left(\frac{5}{7}\right)^{3}}{\frac{7}{5}}=\log_{\left(\frac{7}{5}\right)^{-3}}{\frac{7}{5}}=-\frac{1}{3}\log_{\frac{7}{5}}{\frac{7}{5}}=-\frac{1}{3}\cdot1=-\frac{1}{3}$

→ $\log_{100000}{0,1}=\log_{10^{5}}{10^{-1}}=-\frac{1}{5}\log{10}=-\frac{1}{5}\cdot1=-\frac{1}{5}$

→ $\log_{\frac{1}{5}}{\sqrt{\sqrt[3]{\sqrt[5]{5}}}}=\log_{\frac{1}{5}}{\sqrt[30]{5}}=\log_{5^{-1}}{5^{\frac{1}{30}}}=-\frac{1}{30}\log_{5}{5}=-\frac{1}{30}\cdot1=-\frac{1}{30}$

$E=-\frac{1}{3}-(-\frac{1}{5})-(-\frac{1}{30})=-\frac{1}{3}+\frac{1}{5}+\frac{1}{30}=\frac{-10+6+1}{30}=-\frac{3}{30}=-\frac{1}{10}$

b)

→ $\ln{e}=1$

→ $e^{\ln{\sqrt{2}}}=\sqrt{2}$

→ $\ln{1}=0$

→ $\log{10}=1$

→ $5^{\log_{625}{64}}=5^{\log_{5^{4}}{64}}=5^{\frac{1}{4}\log_{5}{64}}=({5^{\log_{5}{64}}})^{\frac{1}{4}}=64^{\frac{1}{4}}=\sqrt[4]{64}=\sqrt[4]{16\cdot4}=2\cdot\sqrt[4]{4}=2\cdot\sqrt{\sqrt{4}}=2\sqrt{2}$

$E=1-\sqrt{2}+0-1+2\sqrt{2}=\sqrt{2}$