Question

4. Eddie invests £4000 into a savings account, the bank pays him 4% compound

interest per annum. At the end of n years he has £4679.43 to the nearest
penny. What is the value of n?

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\pounds 4679.43\\ P=\textit{original amount deposited}\dotfill &\pounds 4000\\ r=rate\to 4\%\to (4)/(100)\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years \end{cases}`

`4679.43=4000\left(1+(0.04)/(1)\right)^(1\cdot t)\implies \cfrac{4679.43}{4000}=(1.04)^t \\\\\\ \log\left( \cfrac{4679.43}{4000} \right)=\log(1.04^t)\implies \log\left( \cfrac{4679.43}{4000} \right)=t\log(1.04) \\\\\\ \cfrac{\log\left( (4679.43)/(4000) \right)}{\log(1.04)}=t\implies 4\approx t`