Question

Sendo log 2= 0,301 e log 3 = 0,477, calcule:
a) log 12 b) 3,6 d) √216 ?

Answer 1

Olá,

aplique as propriedades,

do produto............$\log(ab)\Rightarrow \log(a)+\log(b)$

do quociente........$\log\left( \dfrac{a}{b}\right)\Rightarrow \log(a)-\log(b)$

da potência..........$\log(b)^n\Rightarrow n\cdot \log(b)$

decorrente da definição (D1).......$\log_b(b)=1$

__________________

$\log(12)=\log(2^2\cdot3)\\ \log(12)=\log(2)^2+\log(3)\\ \log(12)=2\cdot\log(2)+\log(3)\\\\ \log(2)=0,301~~e~~\log(3)=0,477~~lembra??\\\\ \log(12)=2\cdot0,301+0,477\\ \log(12)=0,602+0,477\\\\ \Large\boxed{\log(12)=1,079}$

.................


$\log(3,6)=\log\left( \dfrac{36}{10}\right)\\ \log(3,6)=\log(36)-\log(10)\\\\ mostrando~a~base~oculta..\rightarrow \log(10)=\log_{10}(10)\\\\ \log(3,6)=[\log(2^2\cdot3^2)]-\log_{10}(10)\\ \log(3,6)=[2\cdot\log(2)+2\cdot\log(3)]-\log_{10}(10)\\ \log(3,6)=[2\cdot0,301+2\cdot0,477]-1\\ \log(3,6)=(0,602+0,954)-1\\ \log(3,6)=1,556-1\\\\ \Large\boxed{\log(3,6)=0,556}$

.................


$\log( \sqrt{216})=\log( \sqrt{2^3\cdot3^3})\\ \log( \sqrt{216})=\log( \sqrt{2^3})+\log( \sqrt{3^3})\\\\ \log( \sqrt{216})=\log(2)^{ \tfrac{3}{2}}+\log(3)^{ \tfrac{3}{2} }\\\\ \log( \sqrt{216})= \dfrac{3}{2}\cdot\log(2)+ \dfrac{3}{2}\cdot\log(3)\\\\ \log( \sqrt{216})= \dfrac{3}{2}\cdot(0,301)+ \dfrac{3}{2}\cdot(0,477)\\\\ \log( \sqrt{216})= \dfrac{3\cdot0,301}{2}+ \dfrac{3\cdot0,477}{2}\\\\ \log( \sqrt{216})=0,4515+0,7155\\\\ \Large\boxed{\log( \sqrt{216})=1,167}$


Tenha ótimos estudos ;D