Question

Don cheTD invested $ 6,300 in Account in the

year 20066, Its been reached $8,200 in 2010
Determine the value of the account in 2019

Answer 1

y = abᵗ

we know the initial amount is 6300 on 2006, that's on year 0, namely t = 0.

we also know that it became 8200 on 2010, that's 4 years later, namely t = 4.

what would its value be in 2019, well that's t = 13.

`y = ab^t \\\\[-0.35em] ~\dotfill\\\\ \underset{t=0\hfill }{6300=ab^0}\implies 6300-a(1)\implies 6300=a~\hfill \underline{y=6300b^t} \\\\[-0.35em] ~\dotfill\\\\ \underset{t=4\hfill }{8200=6300b^4}\implies \cfrac{8200}{6300}=b^4\implies \cfrac{82}{63}=b^4\implies \sqrt[4]{\cfrac{82}{63}}=b \\\\\\ ~\hfill \underline{y=6300\left( \sqrt[4]{(82)/(63)} \right)^t} \\\\[-0.35em] ~\dotfill\\\\ \textit{when t = 13}\qquad \qquad y=6300\left( \sqrt[4]{(82)/(63)} \right)^(13)\implies y\approx 14838.1168`