Question

Dados log2=0,3010 log3=0,4771 log 5=0,6990 e log7=0,8450 calcule:
SÓ PRECISO DA A) E) F) G)

Answer 1

Resposta:

A)  log 6²= 2 * [log 3+log 2] = 2* [0,4771+0,301]

E)  log (3²*5) =2*log 3+ log 5 =2*0,4771+0,6990

F)  log 5*2*3=log 5 +log 2+log 3 =0,6990+0,3010+0,4771

G)  log 6/10 =log 6 - log 10 =log 2 +log 3 - log 2-log 5

   = log 3 - log 2 =0,4771 - 0,3010