Question

Considerando log 2 ≅ 0,3, log 3 ≅ 0,48 e log 5 ≅ 0,7, calcule o valor de:

e) log₄ 18

f) log₃₆ 0,5

Answer 1

Utilizando as propriedades dos logaritmos, temos:

$e)~log_418 = log_{2^2}(3^2.2) = \frac{1}{2}log_2(3^2.2) = \\\\\frac{1}{2}.(log_23^2 + log_22) = \frac{1}{2}.(2.log_23+1) = \\\\\frac{1}{2}.(2.\frac{log_3}{log_2} + 1)}= \frac{1}{2}.(2.\frac{0{,}48}{0{,}3} + 1) = \boxed{2{,}1}\\\\\\\\f)~log_{36}0{,}5 = log_{6^2}(2^{-1}) = \frac{-1}{2}.log_62 = \\\\\frac{-1}{2}.\frac{log2}{log6} = \frac{-1}{2}.\frac{0,3}{log2+log3} = \frac{-1}{2}.\frac{0,3}{0,3+0,48} \approx \boxed{-0{,}1923}$