Using compound interest, it is found that the approximate difference in the number of years that calvin and makayla have their money invested is given as follows:
Makayla invests her money 2 years longer.
<h3>What is compound interest?</h3>
The amount of money earned, in compound interest, after t years, is given by:
![A(t) = P\left(1 + \frac{r}{n}\right)^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%5Cleft%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D)
In which:
- A(t) is the amount of money after t years.
- P is the principal(the initial sum of money).
- r is the interest rate(as a decimal value).
- n is the number of times that interest is compounded per year.
- t is the time in years for which the money is invested or borrowed.
For Calvin, the parameters are given as follows:
![P = 400, r = 0.05, n = 12, A(t) = 658.8](https://tex.z-dn.net/?f=P%20%3D%20400%2C%20r%20%3D%200.05%2C%20n%20%3D%2012%2C%20A%28t%29%20%3D%20658.8)
Hence:
![A(t) = P\left(1 + \frac{r}{n}\right)^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%5Cleft%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D)
![658.8= 400\left(1 + \frac{0.05}{12}\right)^{12t}](https://tex.z-dn.net/?f=658.8%3D%20400%5Cleft%281%20%2B%20%5Cfrac%7B0.05%7D%7B12%7D%5Cright%29%5E%7B12t%7D)
![\left(1 + \frac{0.05}{12}\right)^{12t} = \frac{658.8}{400}](https://tex.z-dn.net/?f=%5Cleft%281%20%2B%20%5Cfrac%7B0.05%7D%7B12%7D%5Cright%29%5E%7B12t%7D%20%3D%20%5Cfrac%7B658.8%7D%7B400%7D)
![(1.00416666667)^{12t} = 1.647](https://tex.z-dn.net/?f=%281.00416666667%29%5E%7B12t%7D%20%3D%201.647)
![\log{(1.00416666667)^{12t}} = \log{1.647}](https://tex.z-dn.net/?f=%5Clog%7B%281.00416666667%29%5E%7B12t%7D%7D%20%3D%20%5Clog%7B1.647%7D)
![12t\log{1.00416666667} = \log{1.647}](https://tex.z-dn.net/?f=12t%5Clog%7B1.00416666667%7D%20%3D%20%5Clog%7B1.647%7D)
![t = \frac{\log{1.647}}{12\log{1.00416666667}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B1.647%7D%7D%7B12%5Clog%7B1.00416666667%7D%7D)
![t = 10](https://tex.z-dn.net/?f=t%20%3D%2010)
For Makayla, the parameters are given as follows:
![P = 300, r = 0.06, n = 4, (t) = 613.4](https://tex.z-dn.net/?f=P%20%3D%20300%2C%20r%20%3D%200.06%2C%20n%20%3D%204%2C%20%28t%29%20%3D%20613.4)
Hence:
![A(t) = P\left(1 + \frac{r}{n}\right)^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%5Cleft%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D)
![613.4 = 300\left(1 + \frac{0.06}{4}\right)^{4t}](https://tex.z-dn.net/?f=613.4%20%3D%20300%5Cleft%281%20%2B%20%5Cfrac%7B0.06%7D%7B4%7D%5Cright%29%5E%7B4t%7D)
![\left(1 + \frac{0.06}{4}\right)^{4t} = \frac{613.4}{300}](https://tex.z-dn.net/?f=%5Cleft%281%20%2B%20%5Cfrac%7B0.06%7D%7B4%7D%5Cright%29%5E%7B4t%7D%20%3D%20%5Cfrac%7B613.4%7D%7B300%7D)
![(1.015)^{4t} = 2.0447](https://tex.z-dn.net/?f=%281.015%29%5E%7B4t%7D%20%3D%202.0447)
![\log{(1.015)^{12t}} = \log{2.0447}](https://tex.z-dn.net/?f=%5Clog%7B%281.015%29%5E%7B12t%7D%7D%20%3D%20%5Clog%7B2.0447%7D)
![12t\log{1.015} = \log{2.0447}](https://tex.z-dn.net/?f=12t%5Clog%7B1.015%7D%20%3D%20%5Clog%7B2.0447%7D)
![t = \frac{\log{2.0447}}{4\log{1.015}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B2.0447%7D%7D%7B4%5Clog%7B1.015%7D%7D)
![t = 12](https://tex.z-dn.net/?f=t%20%3D%2012)
12 - 10 = 2, hence Makayla invests her money 2 years longer.
More can be learned about compound interest at brainly.com/question/25781328