Question

Dados log2=0,301 e log3=0,477,o log 7,2 vale:
a) 0,380
b)0,857
c) 0,861
d)1,857
e)1,861

Answer 1

Oi Rafaela,

use as propriedades,

do produto..

$\log(ab)=\log(a)+\log(b)$

do quociente..

$\log\left( \dfrac{a}{b}\right)=\log(a)-\log(b)$

da potência..

$\log(b)^n=n\cdot\log(b)$

e a decorrente da definição (D2)..

$\log_b(b)=1$

_____________


$\log(7,2)= \log\left(\dfrac{72}{10}\right)\\\\ \log(7,2)=\log(72)-\log(10)\\ \log(7,2)=\log(2^3\cdot3^2)-\log_{10}(10)\\ \log(7,2)=[\log(2)^3+\log(3)^2]-\log_{10}(10)\\ \log(7,2)=[3\cdot\log(2)+2\cdot\log(3)]-\log_{10}(10)\\\\ \log(2)=0,301~~;~~\log(3)=0,477~~e~~\log_{10}(10)=1,~~lembra???\\\\ \log(7,2)=[3\cdot0,301+2\cdot0,477]-1\\ \log(7,2)=(0,903+0,954)-1\\ \log(7,2)=1,857-1\\\\ \Large\boxed{\log(7,2)=0,857}$

Ou seja, alternativa B

BJK^^