Question

considerando o log 2=0,301 e log 3=0,477 cauculeo log 25

Answer 1

$log 25 = log 5^{2} = 2.(log 5) = 2.(log \frac{10}{2} ) = 2.(log 10 - log 2) \\ \\ 2.(1-0,301) = 2(0,699) = 1,398$

Espero ter ajudado.