Question

Adotando log 5 7 =1,21 e log 2 0,43

Answer 1

$a) \log_{5}14 = \log_{5}{(2 \cdot 7)} \\ = \log_{5}2 + \log_{5}7 \\ = 0,43 + 1,21 \\ = 1,64 \\ \\ b) \log_{5}{28} = \log_{5}{(2 \cdot 14)} \\ = \log_{5}2 + \log_{5}14 \\ = 0,43 + 1,64 \\ = 2,07$