Answer :
Solution
Part a
We can do the following:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]For this case we can replace and we have:
[tex]A=10000(1+\frac{0.052}{12})^{12\cdot10}=16801.39[/tex]Part b
For this case we can do the following:
[tex]30000=10000(1+\frac{0.052}{12})^{12\cdot t}[/tex]Solving for t we have
[tex]3=(1+\frac{0.052}{12})^{12t}[/tex][tex]\ln (3)=12t\ln (1+\frac{0.052}{12})[/tex][tex]t=\frac{\ln (3)}{12\cdot\ln (1+\frac{0.052}{12})}=21.17[/tex]then the answer is 21.1 years