Respuesta :
Using compound interest, it is found that the approximate difference in the number of years that Calvin and Makayla have their money invested is of 2.
What is compound interest?
The amount of money earned, in compound interest, after t years, is given by:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
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, we have that:
[tex]A(t) = 658.8, P = 400, r = 0.05, n = 12[/tex].
Then:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]658.8 = 400\left(1 + \frac{0.05}{12}\right)^{12t}[/tex]
[tex](1.004167)^{12t} = 1.647[/tex]
[tex]\log{1.004167)^{12t}} = \log{1.647}[/tex]
[tex]12t\log{1.004167} = \log{1.647}[/tex]
[tex]t = \frac{\log{1.647}}{12\log{1.004167}}[/tex]
[tex]t = 10[/tex]
For Makayla, we have that:
[tex]A(t) = 613.04, P = 300, r = 0.06, n = 4[/tex].
Then:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]613.04 = 300\left(1 + \frac{0.06}{4}\right)^{4t}[/tex]
[tex](1.015)^{4t} = 2.0435[/tex]
[tex]\log{(1.015)^{4t}} = \log{2.0435}[/tex]
[tex]4t\log{1.015} = \log{2.0435}[/tex]
[tex]t = \frac{\log{2.0435}}{4\log{1.015}}[/tex]
[tex]t = 12[/tex]
The difference is:
12 - 10 = 2.
The approximate difference in the number of years that Calvin and Makayla have their money invested is of 2.
More can be learned about compound interest at https://brainly.com/question/25781328
