Question

considerando o log2 = 0,301 e log3 = 0,4771 e log5 = 0,699 calcule:
a)log30​
b)logo

c)log ⁷V—2,5 (raiz sétima de 2,5)

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Consequ\hat encias\,da\,de~\!\!finic_{\!\!,}\tilde ao}\\\sf \boxed{1} \log_b1=0\\\sf\boxed{2}\log_bb=1\\\sf\boxed{3} a^{\log_an}=n\\\underline{\rm Consequ\hat encia\,da\,mudanc_{\!\!,}a\,de\,base}\\\sf\log_{b^n}a=\dfrac{1}{\bf n}\cdot\log \sf a\\\\\underline{\rm Propriedades\,operat\acute orias\,dos\,logaritmos}\\\sf \boxed{1}~\log_c(a\cdot b)=\log_ca+\log_cb\\\sf\boxed{2}\log_c\bigg(\dfrac{a}{b}\bigg)=\log_ca-\log_cb\\\\\sf\boxed{3}\log_ba^n=\bf n\cdot\sf\log_ba\end{array}}$

$\large\boxed{\begin{array}{l}\sf \log2=0,301~~\log3=0,4771~~\log5=0,699\\\rm a)~\begin{array}{c|c}\sf30&\sf2\\\sf15&\sf3\\\sf5&\sf5\\\sf1\end{array}\\\sf \log30=\log(2\cdot3\cdot5)\\\sf\log30=\log2+\log3+\log5\\\sf\log30=0,301+0,4771+0,699\\\sf \log30=1,4771\end{array}}$

$\Large\boxed{\begin{array}{l}\rm c)~\sf 2,5=\dfrac{5}{2}\\\\\sf \log\sqrt[\sf7]{\sf2,5}=\log\sqrt[\sf7]{\sf\dfrac{5}{2}}=\log\bigg(\dfrac{5}{2}\bigg)^7\\\sf \log\sqrt[\sf7]{\sf2,5}=7\log\bigg(\dfrac{5}{2}\bigg)\\\\\sf \log\sqrt[\sf7]{\sf2,5}=7\bigg[\log5-\log2\bigg]\\\\\sf\log\sqrt[\sf7]{\sf2,5}=7[0,699-0,301]\\\\\sf \log\sqrt[\sf7]{\sf2,5}=7\cdot0,398\\\sf \log\sqrt[\sf7]{\sf2,5}=2,786\end{array}}$