Question

2) Dados log 2 = 0,3010: log 3 = 0,4771 e log 5 = 0,6990. Calcule a) log 180 b) log 750​

Answer 1

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

$\Large\boxed{\begin{array}{l}\sf \log2=0,3010\\\sf \log3=0,4771\\\sf \log5=0,6990\\\rm a)\\\begin{array}{c|c}\sf180&\sf2\\\sf90&\sf2\\\sf45&\sf3\\\sf15&\sf3\\\sf5&\sf5\\\sf1\end{array}\\\sf \log180=\log(2^2\cdot3^2\cdot5)\\\sf =\log2^2+\log3^2+\log5\\\sf=2\log2+2\log3+\log5\\\sf=2\cdot0,3010+2\cdot0,4771+0,6990\\\sf=0,6020+0,9542+0,6990\\\sf=2,2552\end{array}}$

$\large\boxed{\begin{array}{l}\rm b)\\\begin{array}{c|c}\sf750&\sf2\\\sf375&\sf3\\\sf125&\sf5\\\sf25&\sf5\\\sf5&\sf5\\\sf1\end{array}\\\sf 750=2\cdot3\cdot5^3\\\sf \log750=\log(2\cdot3\cdot5^3)\\\sf=\log2+\log3+\log5^3\\\sf=\log2+\log3+3\log5\\\sf=0,3010+0,4771+3\cdot0,6990\\\sf=0,3010+0,4771+2,097\\\sf=2,8751\end{array}}$