[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{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 \frac{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}[/tex]
[tex]4679.43=4000\left(1+\frac{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( \frac{4679.43}{4000} \right)}{\log(1.04)}=t\implies 4\approx t[/tex]
