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Adrian invested $9,600 in an account paying an interest rate of
of 7 1/2% compounded monthly. Lauren invested $9,600 in an account paying an interest rate of 7 5/8 % compounded continuously. To the nearest dollar, how
much money would Adrian have in his account when Lauren's money has
tripled in value?

Respuesta :

Using compound interest equations, it is found that Lauren will have $626 more in her account.

Compound interest:

The equation for compound interest is:

[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.

Considering continuous compounding, the equation is:

[tex]A(t) = Pe^{rt}[/tex]

For Lauren, we have that:

  • Continuous compounding with a rate of [tex]7\frac{5}{8}\%[/tex], hence [tex]r = 0.07625[/tex]

The time it takes to triple is t for which A(t) = 3P, hence:

[tex]A(t) = Pe^{rt}[/tex]

[tex]3P = Pe^{0.07625t}[/tex]

[tex]e^{0.07625t} = 3[/tex]

[tex]\ln{e^{0.07625t}} = \ln{3}[/tex]

[tex]0.07625t = \ln{3}[/tex]

[tex]t = \frac{\ln{3}}{0.07625}[/tex]

[tex]t = 14.4[/tex]

For Adrian, we have that:

  • Invested $9,600, hence [tex]P = 9600[/tex].
  • Interest rate of [tex]7\frac{1}{2}\%[/tex], hence [tex]r = 0.075[/tex].
  • Compounded monthly, hence [tex]n = 12[/tex].

In 14.4 years, he will have:

[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

[tex]A(14.4) = 9600\left(1 + \frac{0.075}{12}\right)^{12(14.4)}[/tex]

[tex]A(14.4) = 28174[/tex]

Lauren's amount is of 3 x 9600 = 28800.

28800 - 28174 = 626.

Lauren will have $626 more in her account.

To learn more about compound interest equations, you can take a look at https://brainly.com/question/25537936