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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)

Respuesta :

[tex]\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 \frac{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[/tex]