Respuesta :
Answer:
Using continuous interest 6.83 years before she has $1600.
Using continuous compounding, 6.71 years.
Step-by-step explanation:
Compound interest:
The compound interest formula is given by:
[tex]A(t) = P(1 + \frac{r}{n})^{nt}[/tex]
Where 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 unit year and t is the time in years for which the money is invested or borrowed.
Continuous compounding:
The amount of money earned after t years in continuous interest is given by:
[tex]P(t) = P(0)e^{rt}[/tex]
In which P(0) is the initial investment and r is the interest rate, as a decimal.
If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600?
We have to find t for which [tex]A(t) = 1600[/tex] when [tex]P = 1000, r = 0.07, n = 2[/tex]
[tex]A(t) = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]1600 = 1000(1 + \frac{0.07}{2})^{2t}[/tex]
[tex](1.035)^{2t} = \frac{1600}{1000}[/tex]
[tex](1.035)^{2t} = 1.6[/tex]
[tex]\log{1.035)^{2t}} = \log{1.6}[/tex]
[tex]2t\log{1.035} = \log{1.6}[/tex]
[tex]t = \frac{\log{1.6}}{2\log{1.035}}[/tex]
[tex]t = 6.83[/tex]
Using continuous interest 6.83 years before she has $1600
If the compounding is continuous, how long will it be?
We have that [tex]P(0) = 1000, r = 0.07[/tex]
Then
[tex]P(t) = P(0)e^{rt}[/tex]
[tex]1600 = 1000e^{0.07t}[/tex]
[tex]e^{0.07t} = 1.6[/tex]
[tex]\ln{e^{0.07t}} = \ln{1.6}[/tex]
[tex]0.07t = \ln{1.6}[/tex]
[tex]t = \frac{\ln{1.6}}{0.07}[/tex]
[tex]t = 6.71[/tex]
Using continuous compounding, 6.71 years.
