1. Suppose you invest $5010 at an annual interest rate of 6.4% compounded continuously. A.) Write an equation to model this. B) How much money will you have in the account after 2 years?2. Suppose you invest $100 at 6% annual interest. Calculate the amount of money you would have after 1 year if the interest is compounded:A.) quarterly.B.) monthly.C.) daily.3. Suppose you deposit $100,000 in an account today that pays 6% interest compounded annually. Write an equation to determine how long it will take before the amount in your account is $500,000 (Note: you only have to write the equation).I need help because I'm stuck.

To solve our problems, we are going to use the formula for compounded interest: [tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
where
[tex]A[/tex] is the final amount after [tex]t[/tex] years 
[tex]P[/tex] is the initial amount 
[tex]r[/tex] is the interest rate in decimal form 
[tex]n[/tex] is the number of times the interest is compounded per year
[tex]t[/tex] is the time in years

1. A) We know for our problem that the initial investment is $5010, so [tex]P=5000[/tex]. 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%: [tex]r= \frac{6.4}{100} =0.064[/tex]. Since the interest is compounded continuously, it is compounded 365 times per year; therefore, [tex]n=365[/tex]. Lets replace those values in our formula:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]A=5010(1+ \frac{0.064}{365})^{365t} [/tex]

We can conclude that the equation that model this situation is [tex]A=5010(1+ \frac{0.064}{365})^{365t} [/tex]

B. To find the amount of money you will have after 2 years, we are going to replace [tex]t[/tex] with 2 in the equation from point A:
[tex]A=5010(1+ \frac{0.064}{365})^{365t}[/tex]
[tex]A=5010(1+ \frac{0.064}{365})^{(365)(2)}[/tex]
[tex]A=5694.6[/tex]

We can conclude that after 2 years you will have $5694.06 in your account.

2. We know for our problem that [tex]P=100[/tex], [tex]r= \frac{6}{100} =0.06[/tex], and [tex]t=1[/tex].
A. Since the interest is compounded quarterly, it is compounded 4 times per year; therefore, [tex]n=4[/tex]. Lets replace the values in our formula:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]A=100(1+ \frac{0.06}{4} )^{(4)(1)[/tex]
[tex]A=106.14[/tex]
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, [tex]n=12[/tex]. Lets replace the values in our formula:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]A=100(1+ \frac{0.06}{12})^{(12)(1)[/tex]
[tex]A=106.17[/tex]
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, [tex]n=365[/tex]. Lets replace the values in our formula:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]A=100(1+ \frac{0.06}{365})^{(365)(1)} [/tex]
[tex]A=106.18[/tex]
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 [tex]P=100000[/tex]. We also know that the final amount will be $500,000, so [tex]A=500000[/tex]. The interest rate is 6%, so [tex]r= \frac{6}{100} =0.06[/tex]. Since the interest rate is compunded anually, it is compounded 1 time per year; therefore, [tex]n=1[/tex]. Lets replace the values in our formula and solve for [tex]t[/tex]:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]500000=100000(1+ \frac{0.06}{1})^{(1)(t)} [/tex]
[tex]500000=100000(1+0.06)^t[/tex]
[tex] \frac{500000}{100000} =1.06^t[/tex]
[tex]1.06^t=5[/tex]
[tex]ln(1.06^t)=ln(5)[/tex]
[tex]tln(1.06)=ln(5)[/tex]
[tex]t= \frac{ln(5)}{ln(1.06)} [/tex]

We can conclude that [tex]t= \frac{ln(5)}{ln(1.06)} [/tex] is the equation to determine how long it will take before the amount in your account is $500,000

As a bonus: 
[tex]t= \frac{ln(5)}{ln(1.06)} [/tex]
[tex]t=27.6[/tex]
We can conclude that after 27.6 years you will have $500,000 in your account.