Consider the formula,
[tex]A=P(1+\frac{r}{100})^n[/tex]Here, P is the principal, r is the rate of interest, n is the number of periods, and A is the amount.
According to the problem, the compund interest is to be applied quarterly i.e 3 times per year, so the rate of interest is calculated as,
[tex]r=\frac{R}{3}=\frac{3.14}{3}=\frac{157}{150}[/tex]Substitute the values and solve for 'n',
[tex]\begin{gathered} 5341.12=3044(1+\frac{157}{150\times100})^n \\ (1.0105)^n=1.755 \end{gathered}[/tex]Consider the formula,
[tex]\ln (e^x)=e^{\ln x}=x[/tex]Then the equation becomes,
[tex]\begin{gathered} \ln (1.0105)^n=\ln (1.755) \\ n\ln (1.0105)=\ln (1.755) \\ n=\frac{\ln (1.755)}{\ln (1.0105)} \\ n=53.85 \\ n\approx54 \end{gathered}[/tex]Thus, the required number of periods in 54.
The corresponding number of years will be 54 by 3 i.e. 18, since the compounding is done 3 time
