Question

aplicando a definição, calcule o valor dos logaritmos:
F)log25 0,2
G)log2 2√2
H)log2 ∛64
I)√8 4

Answer 1

$F) log 25_{0,2}=log 5^2_{ \frac{1}{5} } =-2log 5_{5} =-2 \\ G)log 2_{2 \sqrt{2} } =log 2_{2*2^ \frac{1}{2} } =log 2_{2^ \frac{3}{2} }= \frac{2}{3}log 2_{2}= \frac{2}{3} \\ H)log 2_{ \sqrt[3]{64} }=log 2_{ \sqrt[3]{2^6} } =log 2_{2^ \frac{6}{3} }= \frac{3}{6} log 2_{2} = \frac{3}{6} = \frac{1}{2} \\ I)log \sqrt{8} _{4} =log \sqrt{2^3} _{2^2} =log 2^{ \frac{3}{2} } _{2^2} = (\frac{3}{2}* \frac{1}{2})log 2_{2} = \frac{3}{4}$