[tex]~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&\$510\\ P=\textit{original amount deposited}\dotfill & \$430\\ r=rate\to 2.5\%\to \frac{2.5}{100}\dotfill &0.025\\ t=years \end{cases}[/tex]
[tex]510=430e^{0.025\cdot t}\implies \cfrac{510}{430}=e^{0.025t}\implies \cfrac{51}{43}=e^{0.025t} \\\\\\ \log_e\left( \cfrac{51}{43} \right)=\log_e\left( e^{0.025t} \right)\implies \ln\left( \cfrac{51}{43} \right)=0.025t \\\\\\ \cfrac{\ln\left( \frac{51}{43} \right)}{0.025}=t\implies 6.83\approx t\implies \stackrel{\textit{rounded up}}{7=t}[/tex]
