# Math Help - Compounding interest problem help

1. ## Compounding interest problem help

Hello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on:

8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337.
a. What is the effect of doubling the amount invested?

b. What is the effect of doubling the annual interest rate?

c. What is the effect of doubling the investment period?

d. Which of the above has the greatest effect on the final amount of the investment?

I can't seem to figure out how to start...

Any suggestions? Thanks!

2. Originally Posted by qcom
Hello everyone, I am working on a math project relating to logarithms, exponential growth and decay, and I have the following problem to work on:

8. The final amount for $5000 invested for 25 years at 10% annual interest compounded semiannually is$57,337.
a. What is the effect of doubling the amount invested?

b. What is the effect of doubling the annual interest rate?

c. What is the effect of doubling the investment period?

d. Which of the above has the greatest effect on the final amount of the investment?

I can't seem to figure out how to start...

Any suggestions? Thanks!
You should know that

$A = P(1 + r)^n$, where P is the principal, r is the percentage interest rate, n is the number of time periods, and A is the final amount.

3. Using Proveit's:
A = 5000(1.05^50) = 57337

25 years, so there are 50 semiannual periods; .10/2 = .05 is rate per period; kapish?

4. Hey thanks a lot 'Prove It' and Wilmer, I think I was mainly confused about the semiannual part, and so you just double the number of years because semiannual means that it is collected twice in one year, right?

Also, when you divide the rate .1 (10% in decimal form) by 2, is that also due to the fact that the interest is collected semiannually?

Otherwise, I know how to proceed and do a-d.

BTW, not much of a concern but I knew the formula as Y = a(1 + r)^t

But it doesn't really make a difference,

5. Correct.
And this is what "happens" during the 25 years:
Code:
         Interest               Balance
0                               5000.00
1          250.00               5250.00
2          262.50               5512.50
3          275.62               5788.12
....
49                             54606.67
50        2730.33              57337.00

6. Alright, now I got it for sure, thanks.

Not sure if you want to help with another problem, but I was also confused with this one, which deals with similar solving techniques, I think...

6. Consider a \$1000 investment that is compounded annually at three different interest rates: 5%, 5.5%, and 6%.

a. Write and graph a function for each interest rate over a time period from 0 to 60 years.

b. Compare the graphs of the three functions.

c. Compare the shapes of the graphs for the first 10 years with the shapes of the graphs between 50 and 60 years.

I think the functions are (for problem 'a'):

y = 1000(1 + .05)^60
y = 1000(1 + .055)^60
y = 1000(1 + .06)^60

Does that look correct?

Now for problem 'b', if I'm not mistaken, just look like horizontal lines waaay up on the y-axis.

One of the graphs, for the first equation I gave, looks like this: http://www.wolframalpha.com/input/?i=graph:+y+%3D+1000(1+%2B+.05)^60

Now for 'c', would I just draw the graphs of the equations but replace the 't' value with 10 giving us new equations:

y = 1000(1 + .05)^10
y = 1000(1 + .055)^10
y = 1000(1 + .06)^10

And then the equations for the last ten years, would we need to find the new 'a' value for our equation, instead of just 1000, and then write a new equation for that?

Just please tell me if I'm way off or something!

7. Originally Posted by qcom
> I think the functions are (for problem 'a'):
> y = 1000(1 + .05)^60
> y = 1000(1 + .055)^60
> y = 1000(1 + .06)^60
> Does that look correct?

Not familiar with graphing programs; but above correct as the FINAL
values (60 years later); seems to me this is really what's needed:
y = 1000(1 + .05)^t where t= 0 to 60

> Now for problem 'b', if I'm not mistaken, just look like horizontal lines
> waaay up on the y-axis.

Well, you'd have t (0 to 60) along x-axis, and the y values along y-axis

> Now for 'c', would I just draw the graphs of the equations but replace
> the 't' value with 10 giving us new equations:
> y = 1000(1 + .05)^10
> y = 1000(1 + .055)^10
> y = 1000(1 + .06)^10

Yes, but: y = 1000(1 + .05)^t where t=1 to 10

> And then the equations for the last ten years, would we need to find
> the new 'a' value for our equation, instead of just 1000, and then write > a new equation for that?

Simply this way: y = 1000(1 + .05)^t where t = 51 to 60
.

8. Ok I think I got you, so we would set up equations like this:

y = 1,000(1.05)^10 y = 1,629
y = 1,000(1.05)^20 y = 2,653
y = 1,000(1.05)^30 y = 4,322
y = 1,000(1.05)^40 y = 7,040
y = 1,000(1.05)^50 y = 11,467
y = 1,000(1.05)^60 y = 18,679

And that's just for one of them and you would do that with the other two as well right?

Alright now I get what you meant by 't' goes along the x-axis as it is basically our 'x' value in these functions, right? And then the 'y' would of course go along the 'y' axis.

Now for c, I think I got that as well.

Does this look good? My only concern is that I may need to do an equation like I did for part 'a' for every single number from 1-60 3 times, because there is 3 different interest rates, or is it fine to only do it in intervals of 10?

9. Originally Posted by qcom
....or is it fine to only do it in intervals of 10?
Dunno. I'd assume it is.
Isn't there a way in whatever graphing program you're using
ti give an instruction, like:
Do for t = 1 to 60 : y = (1 + .05)^t ?