Question

Utilizando as propriedades e considerando que log 2 ≅ 0,3 e log 3 ≅ 0,48, calcule o valor:

a)
$log_{9}(100)$

b)
$log_{36}(0.5)$

c)
$log_{0.25}(144)$

Answer 1

Olá!
a)

$\log_9 100 = \log_9 10^2 = 2\log_9 10$

Aplicando a propriedade da mudança de base:

$2\log_9 10 = 2 \,\dfrac{\log 10}{\log 9} \\\\\ \log 9 = \log 3 \times 3 = \log 3 + \log 3 = 2\log3 \\\\ 2\,\dfrac{1}{2\log3}=\dfrac{1}{\log3} \approx 2,083 \\\\ \log_9 100 \approx 2,083$

b) 

$\log_{36} 0,5 = \log_{36}\dfrac{1}{2} = \log_{36} 1 - \log_{36}2 = -\log_{36}2 \\\\ -\log_{36}2 = - \dfrac{\log2}{\log36} \\\\ \log 36 = \log 4\times4\times2 = 2\log 4 + \log 2 = 1,2 + 0,3 = 1,5 \\\\ -\dfrac{\log2}{\log36} = -\dfrac{0,3}{1,5} = -0,2 \\\\ \log_{36} 0,5 \approx -0,2$

c)

$\log_{0,25}144 = \log_{\tfrac{1}{4}}12^2 = 2\log_{\tfrac{1}{4}}12\\\\ 2\log_{\tfrac{1}{4}}12 = 2\, \dfrac{\log12}{\log 1/4} = 2\, \dfrac{\log3 + \log4}{-\log4} \\\\ 2\, \dfrac{\log3 + \log4}{-\log4} = 2 \times\, -\dfrac{0,48 + 0,6}{0,6} = 2 \times 1,8 = 3,6 \\\\\log_{0,25}144 \approx 3,6$

Espero ter ajudado!