Given:
Principal = $1500
Time = 10 year
Amount after interest compounded continuously = $3935.36
To find:
The rate of interest.
Solution:
The formula for amount after continuous compound interest is:
![A=Pe^(rt)](https://img.qammunity.org/2022/formulas/mathematics/high-school/82sqkrzvl0fpxzvh0rxi0p0g8lcz24clo9.png)
Where, P is principal, r is the rate of interest in decimal and t is time in years.
Putting
in the above formula, we get
![3935.36=1500e^(r(10))](https://img.qammunity.org/2022/formulas/mathematics/high-school/o06i91mrekk1fhhyc2v1thhb8wtel0dwav.png)
![(3935.36)/(1500)=e^(10r)](https://img.qammunity.org/2022/formulas/mathematics/high-school/lejr3x4zw2h10va0linteetpe2rd1om7f8.png)
![2.6236=e^(10r)](https://img.qammunity.org/2022/formulas/mathematics/high-school/11790e8xvvjia2wzhyhwjs528rla5mlbsh.png)
Taking ln on both sides, we get
![\ln(2.6236)=\ln e^(10r)](https://img.qammunity.org/2022/formulas/mathematics/high-school/kdaxmvayb92yj7znfnp5kmy1y46dzcjiiz.png)
![[\because \ln e^x=x]](https://img.qammunity.org/2022/formulas/mathematics/high-school/m24tge2dslw94bp9jx3y240611r7x9yr3r.png)
Divide both sides by 10.
![(0.9645)/(10)=r](https://img.qammunity.org/2022/formulas/mathematics/high-school/jrn09lguuk3ulrr3jxm8a26tc6qi9deoy6.png)
![0.09645=r](https://img.qammunity.org/2022/formulas/mathematics/high-school/un0zndwj1emnc01qfe4cb53bvhbicow3qq.png)
The rate of interest in percentage is:
![r=100* 0.09645](https://img.qammunity.org/2022/formulas/mathematics/high-school/irxjjmz2egeu3j54jfbc03e5qna60ho9c5.png)
![r=9.645\%](https://img.qammunity.org/2022/formulas/mathematics/high-school/yypp7z1phvx2wxp3xwyu4f3pnq70081y0c.png)
Therefore, the required rate of interest is 9.645%.