Answer:
[tex]A= 2400\\P= 1200\\r = 6.5\% = 0.065\\n = 365 \\t = t\\A = P(1 + \frac{r}{n})^{nt}\\\\2400 = 1200(1 +\frac{0.065}{365} )^{365t}\\\\\frac{2400}{1200} = (1.00017808219)^{365t}\\\\(2)^{\frac{1}{365} } = (1.00017808219)^t\\\\\\Taking \ log \ on \ both \ sides\\log(1.00190083768) = t \times log(1.00017808219)\\\\t= \frac{log(1.00190083768)}{log(1.00017808219)} = 10.66\\\\t = 10.7years[/tex]
