Question

You deposit $2500 into an account that pays

3.5% annual interest compounded daily. How long will it take for the
balance to reach $3000

Answer 1

now, this is assuming there are 365 days in 1 year, so

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(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 (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}`

`3000=2500\left(1+(0.035)/(365)\right)^(365\cdot t)\implies \cfrac{3000}{2500}=\left(1+(0.035)/(365)\right)^(365t) \\\\\\ \cfrac{6}{5}=\left(1+(0.035)/(365)\right)^(365t)\implies \cfrac{6}{5}=\left(\cfrac{73007}{73000}\right)^(365t)`

`\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( (6)/(5) \right)~~}{\log\left[ \left((73007)/(73000)\right)^(365) \right]}=t\implies 5.21\approx t\qquad \textit{about 5 years, 2 months and a half}`