To solve our problems, we are going to use the formula for compounded interest:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
where
![A](https://tex.z-dn.net/?f=A)
is the final amount after
![t](https://tex.z-dn.net/?f=t)
years
![P](https://tex.z-dn.net/?f=P)
is the initial amount
![r](https://tex.z-dn.net/?f=r)
is the interest rate in decimal form
![n](https://tex.z-dn.net/?f=n)
is the number of times the interest is compounded per year
![t](https://tex.z-dn.net/?f=t)
is the time in years
1. A) We know for our problem that the initial investment is $5010, so
![P=5000](https://tex.z-dn.net/?f=P%3D5000)
. We also know that the interest rate is 6.4%. To express the interest in decimal form, we are going to divide it by 100%:
![r= \frac{6.4}{100} =0.064](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B6.4%7D%7B100%7D%20%3D0.064)
. Since the interest is compounded continuously, it is compounded 365 times per year; therefore,
![n=365](https://tex.z-dn.net/?f=n%3D365)
. Lets replace those values in our formula:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
![A=5010(1+ \frac{0.064}{365})^{365t}](https://tex.z-dn.net/?f=A%3D5010%281%2B%20%5Cfrac%7B0.064%7D%7B365%7D%29%5E%7B365t%7D%20)
We can conclude that the equation that model this situation is
B. To find the amount of money you will have after 2 years, we are going to replace
![t](https://tex.z-dn.net/?f=t)
with 2 in the equation from point
A:
![A=5010(1+ \frac{0.064}{365})^{365t}](https://tex.z-dn.net/?f=A%3D5010%281%2B%20%5Cfrac%7B0.064%7D%7B365%7D%29%5E%7B365t%7D)
![A=5010(1+ \frac{0.064}{365})^{(365)(2)}](https://tex.z-dn.net/?f=A%3D5010%281%2B%20%5Cfrac%7B0.064%7D%7B365%7D%29%5E%7B%28365%29%282%29%7D)
![A=5694.6](https://tex.z-dn.net/?f=A%3D5694.6)
We can conclude that after 2 years you will have $5694.06 in your account.
2. We know for our problem that
![P=100](https://tex.z-dn.net/?f=P%3D100)
,
![r= \frac{6}{100} =0.06](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B6%7D%7B100%7D%20%3D0.06)
, and
![t=1](https://tex.z-dn.net/?f=t%3D1)
.
A. Since the interest is compounded quarterly, it is compounded 4 times per year; therefore,
![n=4](https://tex.z-dn.net/?f=n%3D4)
. Lets replace the values in our formula:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
![A=100(1+ \frac{0.06}{4} )^{(4)(1)](https://tex.z-dn.net/?f=A%3D100%281%2B%20%5Cfrac%7B0.06%7D%7B4%7D%20%29%5E%7B%284%29%281%29)
![A=106.14](https://tex.z-dn.net/?f=A%3D106.14)
We can conclude that after a year you will have $106.14 in your account.
B. Since the interest is compounded monthly, it is compounded 12 times per year; therefore,
![n=12](https://tex.z-dn.net/?f=n%3D12)
. Lets replace the values in our formula:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
![A=100(1+ \frac{0.06}{12})^{(12)(1)](https://tex.z-dn.net/?f=A%3D100%281%2B%20%5Cfrac%7B0.06%7D%7B12%7D%29%5E%7B%2812%29%281%29)
![A=106.17](https://tex.z-dn.net/?f=A%3D106.17)
We can conclude that after a year you will have $106.17 in your account.
C. Since the interest is compounded daily, it is compounded 365 times per year; therefore,
![n=365](https://tex.z-dn.net/?f=n%3D365)
. Lets replace the values in our formula:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
![A=100(1+ \frac{0.06}{365})^{(365)(1)}](https://tex.z-dn.net/?f=A%3D100%281%2B%20%5Cfrac%7B0.06%7D%7B365%7D%29%5E%7B%28365%29%281%29%7D%20)
![A=106.18](https://tex.z-dn.net/?f=A%3D106.18)
We can conclude that after a year you will have $106.18 in your account.
3. We know for our problem that the initial investment is $100,000, so
![P=100000](https://tex.z-dn.net/?f=P%3D100000)
. We also know that the final amount will be $500,000, so
![A=500000](https://tex.z-dn.net/?f=A%3D500000)
. The interest rate is 6%, so
![r= \frac{6}{100} =0.06](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B6%7D%7B100%7D%20%3D0.06)
. Since the interest rate is compunded anually, it is compounded 1 time per year; therefore,
![n=1](https://tex.z-dn.net/?f=n%3D1)
. Lets replace the values in our formula and solve for
![t](https://tex.z-dn.net/?f=t)
:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
![500000=100000(1+ \frac{0.06}{1})^{(1)(t)}](https://tex.z-dn.net/?f=500000%3D100000%281%2B%20%5Cfrac%7B0.06%7D%7B1%7D%29%5E%7B%281%29%28t%29%7D%20)
![500000=100000(1+0.06)^t](https://tex.z-dn.net/?f=500000%3D100000%281%2B0.06%29%5Et)
![\frac{500000}{100000} =1.06^t](https://tex.z-dn.net/?f=%20%5Cfrac%7B500000%7D%7B100000%7D%20%3D1.06%5Et)
![1.06^t=5](https://tex.z-dn.net/?f=1.06%5Et%3D5)
![ln(1.06^t)=ln(5)](https://tex.z-dn.net/?f=ln%281.06%5Et%29%3Dln%285%29)
![tln(1.06)=ln(5)](https://tex.z-dn.net/?f=tln%281.06%29%3Dln%285%29)
![t= \frac{ln(5)}{ln(1.06)}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bln%285%29%7D%7Bln%281.06%29%7D%20)
We can conclude that
![t= \frac{ln(5)}{ln(1.06)}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bln%285%29%7D%7Bln%281.06%29%7D%20)
is the equation to determine how long it will take before the amount in your account is $500,000
As a bonus:
![t= \frac{ln(5)}{ln(1.06)}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bln%285%29%7D%7Bln%281.06%29%7D%20)
![t=27.6](https://tex.z-dn.net/?f=t%3D27.6)
We can conclude that after 27.6 years you will have $500,000 in your account.