Question

Log 2 = 0,30; Log 3 = 0,47; Log 5=0,69.
A- Log 4
B- Log 12
C- Log √2

Answer 1

A) Log 4
Log 2*2
Log 2 + Log 2
0,3 + 0,3 = 0,6

B) Log 12
Log 2*2*3
Log 2 + Log 2 + Log 3
0,3 + 0,3 + 0,47 = 1,07

C) Log √2
Log 2^1/2
1/2*Log 2
1/2*0,3 = 0,15
 

Answer 2

Utilizaremos as propriedades operatórias dos logaritmos
(para uma mesma base $b$ onde $b>1$ ou $0<b<1$)

$\bullet\;\;(i)$ Logaritmo do produto é a soma dos logaritmos:

$\mathrm{\ell og}_{b\,}(p\cdot q)=\mathrm{\ell og}_{b\,}p+\mathrm{\ell og}_{b\,}q$


$\bullet\;\;(ii)$ Logaritmo de uma potência

$\mathrm{\ell og}_{b\,}(p^{k})=k\cdot \mathrm{\ell og}_{b\,}p$


A questão informa que

$\mathrm{\ell og\,}2=0,30;\;\;\mathrm{\ell og\,}3=0,47;\;\;\mathrm{\ell og\,}5=0,69.$


Aplicando as propriedades mencionadas, temos

A) $\mathrm{\ell og}\,4$

$\mathrm{\ell og\,}4=\mathrm{\ell og\,}(2^{2})\\ \\ \mathrm{\ell og\,}4=2\cdot \mathrm{\ell og\,}2 \\ \\ \mathrm{\ell og\,}4=2\cdot 0,30\\ \\ \mathrm{\ell og\,}4=0,60$


B) $\mathrm{\ell og}\,12$

$\mathrm{\ell og}\,12=\mathrm{\ell og}(2^{2}\cdot 3)\\ \\ \mathrm{\ell og}\,12=\mathrm{\ell og}(2^{2})+\mathrm{\ell og\,}3\\ \\ \mathrm{\ell og}\,12=2\cdot \mathrm{\ell og\,}2+\mathrm{\ell og\,}3\\ \\ \mathrm{\ell og}\,12=2\cdot 0,30+0,47\\ \\ \mathrm{\ell og}\,12=0,60+0,47\\ \\ \mathrm{\ell og}\,12=1,07$


C) $\mathrm{\ell og}\,\sqrt{2}$

$\mathrm{\ell og}\,\sqrt{2}=\mathrm{\ell og\,}(2^{1/2})\\ \\ \mathrm{\ell og}\,\sqrt{2}=\dfrac{1}{2}\cdot \mathrm{\ell og\,}2\\ \\ \mathrm{\ell og}\,\sqrt{2}=\dfrac{1}{2}\cdot 0,30\\ \\ \mathrm{\ell og}\,\sqrt{2}=0,15$