José que saber quantas prestações iguais e consecutivas de R$ 6.500,00 deverá aplicar mensalmente, para que obtenha um montante de R$ 206.000,00, sendo que á taxa aplicada é de 3% ao mês. Em que prazo João atingirá seu objetivo?
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Bom dia!
Usando a equação abaixo chegamos a:
![\displaystyle{FV=PMT\cdot\left[\frac{\left(1+i\right)^{n}-1}{i}\right]}\\\displaystyle{206\,000=6\,500\cdot\left[\frac{\left(1+3\%\right)^{n}-1}{3\%}\right]}\\\displaystyle{\frac{206\,000}{6\,500}=\frac{\left(1+0,03\right)^{n}-1}{0,03}}\\\displaystyle{\frac{0,03\cdot{206\,000}}{6\,500}=1,03^{n}-1}\\\displaystyle{1,03^{n}=\frac{0,03\cdot{206\,000}}{6\,500}+1}\\\displaystyle{n=\frac{\log\left(\frac{0,03\cdot{206\,000}}{6\,500}+1\right)}{\log{1,3}}}\\\displaystyle{\boxed{n\approx{23}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7BFV%3DPMT%5Ccdot%5Cleft%5B%5Cfrac%7B%5Cleft%281%2Bi%5Cright%29%5E%7Bn%7D-1%7D%7Bi%7D%5Cright%5D%7D%5C%5C%5Cdisplaystyle%7B206%5C%2C000%3D6%5C%2C500%5Ccdot%5Cleft%5B%5Cfrac%7B%5Cleft%281%2B3%5C%25%5Cright%29%5E%7Bn%7D-1%7D%7B3%5C%25%7D%5Cright%5D%7D%5C%5C%5Cdisplaystyle%7B%5Cfrac%7B206%5C%2C000%7D%7B6%5C%2C500%7D%3D%5Cfrac%7B%5Cleft%281%2B0%2C03%5Cright%29%5E%7Bn%7D-1%7D%7B0%2C03%7D%7D%5C%5C%5Cdisplaystyle%7B%5Cfrac%7B0%2C03%5Ccdot%7B206%5C%2C000%7D%7D%7B6%5C%2C500%7D%3D1%2C03%5E%7Bn%7D-1%7D%5C%5C%5Cdisplaystyle%7B1%2C03%5E%7Bn%7D%3D%5Cfrac%7B0%2C03%5Ccdot%7B206%5C%2C000%7D%7D%7B6%5C%2C500%7D%2B1%7D%5C%5C%5Cdisplaystyle%7Bn%3D%5Cfrac%7B%5Clog%5Cleft%28%5Cfrac%7B0%2C03%5Ccdot%7B206%5C%2C000%7D%7D%7B6%5C%2C500%7D%2B1%5Cright%29%7D%7B%5Clog%7B1%2C3%7D%7D%7D%5C%5C%5Cdisplaystyle%7B%5Cboxed%7Bn%5Capprox%7B23%7D%7D%7D "\displaystyle{FV=PMT\cdot\left[\frac{\left(1+i\right)^{n}-1}{i}\right]}\\\displaystyle{206\,000=6\,500\cdot\left[\frac{\left(1+3\%\right)^{n}-1}{3\%}\right]}\\\displaystyle{\frac{206\,000}{6\,500}=\frac{\left(1+0,03\right)^{n}-1}{0,03}}\\\displaystyle{\frac{0,03\cdot{206\,000}}{6\,500}=1,03^{n}-1}\\\displaystyle{1,03^{n}=\frac{0,03\cdot{206\,000}}{6\,500}+1}\\\displaystyle{n=\frac{\log\left(\frac{0,03\cdot{206\,000}}{6\,500}+1\right)}{\log{1,3}}}\\\displaystyle{\boxed{n\approx{23}}}")
Espero ter ajudado!
Usando a equação abaixo chegamos a:
Espero ter ajudado!
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