Alguém me ajuda?
P(T) = 1,6 . 10 ^bT
Determine b sabendo que:
T é a temperatura em °C
P(T) é o percentual.
P(55) = 20
dados: log 2= 0,3
log 3= 0,48
log 5 = 0,7
munirdaud:
o T está elevado com o b, né?
Soluções para a tarefa
Respondido por
2
Basta aplicar log na equação:






![(2*log2)*5 = (4*log2)+([55b-1*]log10) (2*log2)*5 = (4*log2)+([55b-1*]log10)](https://tex.z-dn.net/?f=%282%2Alog2%29%2A5+%3D+%284%2Alog2%29%2B%28%5B55b-1%2A%5Dlog10%29)
![(2*log2)+log5 = (4*0,3)+([55b-1]*1) (2*log2)+log5 = (4*0,3)+([55b-1]*1)](https://tex.z-dn.net/?f=%282%2Alog2%29%2Blog5+%3D+%284%2A0%2C3%29%2B%28%5B55b-1%5D%2A1%29)








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