Question

Sabendo que log de 2=0,30 e log de 3=0,47 calcule utilizando as propriedades operatórias:
A) log de 1,8 na base 2
B) log de 5 na base 2
C) log de b na base 3.2

Se possível responda toda por favor. E precisa de esclarecimento

Answer 1

Vou listar as principais propriedades operatórias dos logaritmos. Elas serão utilizadas para resolver as expressões:

$\begin{array}{rclc} \mathrm{\ell og}_{b}\left(x\cdot y \right )&=&\mathrm{\ell og}_{b}\,x+\mathrm{\ell og}_{b}\,y&\;\;\;\text{(i)}\\ \\ \mathrm{\ell og}_{b}\left(\dfrac{x}{y} \right )&=&\mathrm{\ell og}_{b}\,x-\mathrm{\ell og}_{b}\,y&\;\;\;\text{(ii)}\\ \\ \mathrm{\ell og}_{b}\left(x^{n} \right )&=&n\cdot \mathrm{\ell og}_{b}\,x&\;\;\;\text{(iii)}\\ \\ \mathrm{\ell og}_{b}\,x&=&\dfrac{\mathrm{\ell og}_{c}\,x}{\mathrm{\ell og}_{c}\,b}&\;\;\;\text{(iv)} \end{array}$


A propriedade $\text{(iv)}$ também é conhecida como lei de mudança de base (da base $b$ para a base $c$).


Sendo assim

A) $\mathrm{\ell og}_{2\,}1,8$

$=\dfrac{\mathrm{\ell og\,}1,8}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og}\left(^{18}\!\!\!\diagup\!\!_{10} \right )}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og\,}18-\mathrm{\ell og\,}10}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og}\left(2\cdot 3^{2} \right )-\mathrm{\ell og\,}10}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og\,}2+\mathrm{\ell og}\left(3^{2} \right )-\mathrm{\ell og\,}10}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og\,}2+2\cdot\mathrm{\ell og\,}3-\mathrm{\ell og\,}10}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\left(0,30 \right )+2\cdot \left(0,47 \right )-1}{\left(0,30 \right )}\\ \\ =\dfrac{0,30+0,94-1}{0,30}\\ \\ =\dfrac{0,24}{0,30}\\ \\ =0,80\\ \\ \\ \boxed{\mathrm{\ell og}_{2\,}1,8=0,80}$


B) $\mathrm{\ell og}_{2\,}5$

$=\dfrac{\mathrm{\ell og\,}5}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og}\left(^{10}\!\!\!\diagup\!\!_{2}\right)}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{\ell og\,}10-\mathrm{\ell og\,}2}{\mathrm{\ell og\,}2}\\ \\ =\dfrac{\mathrm{1-\left(0,30 \right )}}{\left(0,30 \right )}\\ \\ =\dfrac{0,70}{0,30}\\ \\ =2,33\\ \\ \\ \boxed{\mathrm{\ell og}_{2\,}5=2,33}$


C) $\mathrm{\ell og}_{3,2\,}b$

$=\dfrac{\mathrm{\ell og\,}b}{\mathrm{\ell og\,}3,2}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{\mathrm{\ell og}\left(^{32}\!\!\diagup\!\!_{10}\right)}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{\mathrm{\ell og\,}32-\mathrm{\ell og\,}10}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{\mathrm{\ell og}\left(2^{5} \right )-\mathrm{\ell og\,}10}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{5\cdot \mathrm{\ell og\,}2-\mathrm{\ell og\,}10}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{5\cdot \left(0,30 \right )-1}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{1,50-1}\\ \\ =\dfrac{\mathrm{\ell og\,}b}{0,50}\\ \\ =2\,\mathrm{\ell og\,}b\\ \\ \\ \boxed{\mathrm{\ell og}_{3,2\,}b=2\,\mathrm{\ell og\,}b}$