Páginas: 2 (290 palavras) Publicado: 6 de junho de 2011
100.000*(1+8,8/100)^M = 400.000*(1-(1/5))^M
(1+8,8/100)^M/(1-(1/5))^M = 400.000/100.000(1+8,8/100)^M/(1-(20/100))^M = 4
((100+8,8)/100)^M/((100-20/100)^M = 4
(108,8/100)^M/(80/100)^M = 4
(108,8/80)^M = 41,36^M = 2^2
log 1,36^M = log 2^2
M*log 1,36 = 2*log 2
M = 2*log 2/log 1,36
M = 2*0,301/0,1335
M =4,51 meses

M = 5 meses
(1+0,088)^M/(1-3)^M = 400.000/100.000
(1,088)^M/(-2)^M = 4
(-1,088/2)^M = 4(-0,544)^M = 2^2
log (-0,544)^M = log 2^2
(1+0,088)^M/(1-3)^M = 400.000/100.000
(1,088)^M/(-2)^M = 4(-1,088/2)^M = 4
(-0,544)^M = 2^2
log (-0,544)^M = log 2^2
M*log (-0,544) = 2*log 2
M = 2*log 2/log(-0,544)
M = 2*0,301/log (-0,544)

Esquisito, né?

Ou será que isso?

100.000*(1+8,8/100)^M =400.000*(1-(1/5))^M
(1+8,8/100)^M/(1-(1/5))^M = 400.000/100.000
(1+8,8/100)^M/(1-(20/100))^M = 4((100+8,8)/100)^M/((100-20/100)^M = 4
(108,8/100)^M/(80/100)^M = 4
(108,8/80)^M = 4
1,36^M = 2^2
log 1,36^M =log 2^2
M*log 1,36 = 2*log 2
M = 2*log 2/log 1,36
M = 2*0,301/0,1335
M = 4,51 meses

M = 5 meses

Por favor, assinar para o acesso.