We have been given that Mr.Wilkinson invests £3000 at a compound interest rate of 2.2% per annum.
We will use compound interest formula to answer our problem.
[tex]A=P\cdot(1+\frac{r}{n} )^{nT}[/tex], where P= principle amount, A=Amount after T years, n=Period of compounding, r= interest rate (decimal).
A in our case will be [tex]3000+ 800= 3800[/tex]
Upon substituting our given values in above formula we will get,
[tex]3800=3000\cdot(1+\frac{0.022}{1} )^{1*T}[/tex]
[tex]3800=3000\cdot(1.022)^{*T}[/tex]
Upon taking natural log of both sides of our equation we will get,
[tex]\text{ln} 3800=\text{ln}3000\cdot(1.022)^{*T}[/tex]
[tex]\text{ln} 3800=T\cdot \text{ln}3000\cdot(1.022)[/tex]
[tex]T=\frac{\text{ln }3800}{ \text{ln }3000(1.022)}[/tex]
[tex]T=1.026734[/tex]
Therefore, Mr. Wilkinson will have to invest his money for at least 2 years to earn more than £800 interest.
