Respuesta :
Using continuous compounding and compound interest, it is found that it takes 5.04 years longer for Ella's money to double than for Penelope's money to double.
The amount of money, after t years, using continuous compounding, is given by:
[tex]A(t) = A(0)e^{rt}[/tex]
In which:
- A(0) is the initial amount.
- r is the interest rate, as a decimal.
In this problem, Penelope's interest rate, as a percentage, is:
[tex]1\frac{3}{8}\% = 1 + \frac{3}{8} = 1.375\%[/tex]
Hence [tex]r = 0.01375[/tex]
The time to double is t for which A(t) = 2A(0), then:
[tex]A(t) = A(0)e^{rt}[/tex]
[tex]2A(0) = A(0)e^{0.01375t}[/tex]
[tex]e^{0.01375t} = 2[/tex]
[tex]\ln{e^{0.01375t}} = \ln{2}[/tex]
[tex]0.01375t = \ln{2}[/tex]
[tex]t = \frac{\ln{2}}{0.01375}[/tex]
[tex]t = 50.41[/tex]
It takes 50.41 years for Penelope's amount to double.
Compound interest:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
- 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 Ella's, we have that:
- Compounded daily, hence [tex]n = 365[/tex].
- Rate, as a percent, of [tex]1\frac{1}{4} = 1 + \frac{1}{4} = 1.25\%[/tex], hence [tex]r = 0.0125[/tex]
The time to double is t for which A(t) = 2P, hence:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]2P = P\left(1 + \frac{0.0125}{365}\right)^{365t}[/tex]
[tex](1.00003424658)^{365t} = 2[/tex]
[tex]\log{(1.00003424658)^{365t}} = \log{2}[/tex]
[tex]365t\log{1.00003424658} = \log{2}[/tex]
[tex]t = \frac{\log{2}}{365\log{1.00003424658}}[/tex]
[tex]t = 55.45[/tex]
It takes 55.45 years for Ella's amount to double.
55.45 - 50.41 = 5.04
It takes 5.04 years longer for Ella's money to double than for Penelope's money to double.
A similar problem is given at https://brainly.com/question/24507395
