calcular a) log ₈32 b)log₁₊₄128 c)log₉27 D)log₁₊₃3
Soluções para a tarefa
Respondido por
0
Ola Anderson
a)
log8(32) = log(32)/log(8) = log(2^5)/log(2^3) = 5/3
b)
log5(128) = log(128)/log(5) = log(2^7)/log(5)
log(5) = log(10/2) = log(10) - log(2) = 1 - log(2)
log(2^7)/log(5) = 7*log(2)/(1 - log(2))
c)
log4(3) = log(3)/log(4) = log(3)/2log(2)
a)
log8(32) = log(32)/log(8) = log(2^5)/log(2^3) = 5/3
b)
log5(128) = log(128)/log(5) = log(2^7)/log(5)
log(5) = log(10/2) = log(10) - log(2) = 1 - log(2)
log(2^7)/log(5) = 7*log(2)/(1 - log(2))
c)
log4(3) = log(3)/log(4) = log(3)/2log(2)
andersonps32013:
falto a d amigo
Perguntas interessantes