now, this is assuming there are 365 days in 1 year, so
[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$3000\\ P=\textit{original amount deposited}\dotfill &\$2500\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years \end{cases}[/tex]
[tex]3000=2500\left(1+\frac{0.035}{365}\right)^{365\cdot t}\implies \cfrac{3000}{2500}=\left(1+\frac{0.035}{365}\right)^{365t} \\\\\\ \cfrac{6}{5}=\left(1+\frac{0.035}{365}\right)^{365t}\implies \cfrac{6}{5}=\left(\cfrac{73007}{73000}\right)^{365t}[/tex]
[tex]\log\left( \cfrac{6}{5} \right)=\log\left[ \left(\cfrac{73007}{73000}\right)^{365t} \right]\implies \log\left( \cfrac{6}{5} \right)=t\log\left[ \left(\cfrac{73007}{73000}\right)^{365} \right] \\\\\\ \cfrac{~~ \log\left( \frac{6}{5} \right)~~}{\log\left[ \left(\frac{73007}{73000}\right)^{365} \right]}=t\implies 5.21\approx t\qquad \textit{about 5 years, 2 months and a half}[/tex]
