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:
![A(t) = P(1 + \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D)
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:
![P(t) = P(0)e^{rt}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29e%5E%7Brt%7D)
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
when ![P = 1000, r = 0.07, n = 2](https://tex.z-dn.net/?f=P%20%3D%201000%2C%20r%20%3D%200.07%2C%20n%20%3D%202)
![A(t) = P(1 + \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%28t%29%20%3D%20P%281%20%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D)
![1600 = 1000(1 + \frac{0.07}{2})^{2t}](https://tex.z-dn.net/?f=1600%20%3D%201000%281%20%2B%20%5Cfrac%7B0.07%7D%7B2%7D%29%5E%7B2t%7D)
![(1.035)^{2t} = \frac{1600}{1000}](https://tex.z-dn.net/?f=%281.035%29%5E%7B2t%7D%20%3D%20%5Cfrac%7B1600%7D%7B1000%7D)
![(1.035)^{2t} = 1.6](https://tex.z-dn.net/?f=%281.035%29%5E%7B2t%7D%20%3D%201.6)
![\log{1.035)^{2t}} = \log{1.6}](https://tex.z-dn.net/?f=%5Clog%7B1.035%29%5E%7B2t%7D%7D%20%3D%20%5Clog%7B1.6%7D)
![2t\log{1.035} = \log{1.6}](https://tex.z-dn.net/?f=2t%5Clog%7B1.035%7D%20%3D%20%5Clog%7B1.6%7D)
![t = \frac{\log{1.6}}{2\log{1.035}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clog%7B1.6%7D%7D%7B2%5Clog%7B1.035%7D%7D)
![t = 6.83](https://tex.z-dn.net/?f=t%20%3D%206.83)
Using continuous interest 6.83 years before she has $1600
If the compounding is continuous, how long will it be?
We have that ![P(0) = 1000, r = 0.07](https://tex.z-dn.net/?f=P%280%29%20%3D%201000%2C%20r%20%3D%200.07)
Then
![P(t) = P(0)e^{rt}](https://tex.z-dn.net/?f=P%28t%29%20%3D%20P%280%29e%5E%7Brt%7D)
![1600 = 1000e^{0.07t}](https://tex.z-dn.net/?f=1600%20%3D%201000e%5E%7B0.07t%7D)
![e^{0.07t} = 1.6](https://tex.z-dn.net/?f=e%5E%7B0.07t%7D%20%3D%201.6)
![\ln{e^{0.07t}} = \ln{1.6}](https://tex.z-dn.net/?f=%5Cln%7Be%5E%7B0.07t%7D%7D%20%3D%20%5Cln%7B1.6%7D)
![0.07t = \ln{1.6}](https://tex.z-dn.net/?f=0.07t%20%3D%20%5Cln%7B1.6%7D)
![t = \frac{\ln{1.6}}{0.07}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Cln%7B1.6%7D%7D%7B0.07%7D)
![t = 6.71](https://tex.z-dn.net/?f=t%20%3D%206.71)
Using continuous compounding, 6.71 years.