Em qtos meses a diferença entre os montantes de duas aplicações as taxas de 10 e 18% a.a , será igual a 24% do valor da aplicação inicial
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Vou chamar os montantes resultantes das taxas de 18 e 10%, respectivamente, de M1 e M2. Formulando matematicamente o enunciado temos:
![M_{1} - M_{2}=0,24C
[C(1+0,18)^{t}]-[C(1+0,1)^{t}]=0,24C
[(1,18)^{t}C]-[(1,1)^{t}C]=0,24C
(1,18-1,1)^{t}C=0,24C
(0,08)^{t}C=0,24C
(8.10^{-2})^{t}=3.(8.10^{-2})
Log(8.10^{-2})^{t}=Log3.(8.10^{-2})
tLog8.10^{-2}=Log3+Log(8.10^{-2})](https://tex.z-dn.net/?f=M_%7B1%7D+-+M_%7B2%7D%3D0%2C24C%0A%0A%5BC%281%2B0%2C18%29%5E%7Bt%7D%5D-%5BC%281%2B0%2C1%29%5E%7Bt%7D%5D%3D0%2C24C%0A%0A%5B%281%2C18%29%5E%7Bt%7DC%5D-%5B%281%2C1%29%5E%7Bt%7DC%5D%3D0%2C24C%0A%0A%281%2C18-1%2C1%29%5E%7Bt%7DC%3D0%2C24C%0A%0A%280%2C08%29%5E%7Bt%7DC%3D0%2C24C%0A%0A%288.10%5E%7B-2%7D%29%5E%7Bt%7D%3D3.%288.10%5E%7B-2%7D%29%0A%0ALog%288.10%5E%7B-2%7D%29%5E%7Bt%7D%3DLog3.%288.10%5E%7B-2%7D%29%0A%0AtLog8.10%5E%7B-2%7D%3DLog3%2BLog%288.10%5E%7B-2%7D%29+ "M_{1} - M_{2}=0,24C
[C(1+0,18)^{t}]-[C(1+0,1)^{t}]=0,24C
[(1,18)^{t}C]-[(1,1)^{t}C]=0,24C
(1,18-1,1)^{t}C=0,24C
(0,08)^{t}C=0,24C
(8.10^{-2})^{t}=3.(8.10^{-2})
Log(8.10^{-2})^{t}=Log3.(8.10^{-2})
tLog8.10^{-2}=Log3+Log(8.10^{-2})")
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