Given:
Principal value = $65,550
Rate of interest = 3% compounded every six month
To find:
The taken to have $100,000 in Jack's account.
Solution:
The formula for amount is
![A=P(1+(r)/(n))^(nt)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/idqim6yta24jxbv01m2j0bx5wa3qd5n4hr.png)
where, P is principal, r is rate of interest, n is number of times interest compounded in an year, t is time in number of years.
Interest compounded every six month. It means, interest compounded 2 times in an year.
Substitute A=100000, r=0.03 and n=2 in the above formula.
![100000=65550(1+(0.03)/(2))^(2t)](https://img.qammunity.org/2021/formulas/mathematics/college/tli1qsadty3yu2ylf7micjgc2td7lfoank.png)
![(100000)/(65550)=(1+0.015)^(2t)](https://img.qammunity.org/2021/formulas/mathematics/college/x0pvh9qf0x04yx5ivaw9dm4zy2ayqjdcs2.png)
![1.525553=(1.015)^(2t)](https://img.qammunity.org/2021/formulas/mathematics/college/rxpy96vsf6agrqlhxw1l2qa4mtlnhb70lc.png)
Taking log on both sides.
![\log (1.525553)=\log (1.015)^(2t)](https://img.qammunity.org/2021/formulas/mathematics/college/a39dkqwnrrhes1nhe968ogdrgfanhgf5hh.png)
![\log (1.525553)=2t\log (1.015)](https://img.qammunity.org/2021/formulas/mathematics/college/vnf8dklzxo4tmdfcutrbxkhgwesjcjirz5.png)
![(\log (1.525553))/(2\log (1.015))=t](https://img.qammunity.org/2021/formulas/mathematics/college/w5bwoqt7e7ru23sxt0oc14dpmqqqxuepki.png)
![t=14.1838928](https://img.qammunity.org/2021/formulas/mathematics/college/we5z9hk239jel2b0ywpv7de852hbd33n4x.png)
![t\approx 14.18](https://img.qammunity.org/2021/formulas/mathematics/college/c25iga7g2mr4dz1e4qxzplbzmun5itosuf.png)
Therefore, after 14.18 year the amount will reach at $100,000 or we can say that in 15th year the amount will reach at $100,000.