Question

How much money should be invested today in an account that earns 3.5%, compound daily, in order to accumulate $75000 in 10 years (assume n=365)

Answer 1

`\bf ~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill &\$75000\\ P=\textit{original amount deposited}\\ r=rate\to 3.5\%\to (3.5)/(100)\dotfill &0.035\\ t=years\dotfill &10 \end{cases} \\\\\\ 75000=Pe^(0.035\cdot 10)\implies 75000=Pe^(0.35)\implies \cfrac{75000}{e^(0.35)}=P \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 52851.61\approx P~\hfill`