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### Compound interest to a specific amount

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Date: 07/24/97 at 13:24:59
From: Patrick Burke
Subject: Compound interest to a specific amount

Hi,

I've looked at your FAQ, and found dozens of compound interest
formulae and explanations, but I'm sort of math-impaired and haven't
been able to use them to find what I want.

I'm not interested in the final amount, I'm interested in the time it
would take for a regular deposit account to equal one million dollars,
assuming monthly payments of \$100.00 at, say, 8, 10 and 12 percent
interest.

The problem for me is t, time, is an exponential function in the
compound interest formulae, so I have been ending up with the mt-1
root of \$100,00, and things like that, and I don't know how to take
the n-1 root of some number, so I get stuck.

Here is my question:

If I have a brokerage account, and I deposit \$100 into it every month,
how long will it take that account to reach \$1,000,000 at:

8 percent
10 percent
12 percent

assuming interest compounded monthly?

Thanks,

Patrick Burke
Training Specialist
The University of Chicago
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Date: 07/24/97 at 14:20:33
From: Doctor Rob
Subject: Re: Compound interest to a specific amount

Patrick,

Let i be the monthly interest rate.  This is 1/12 the yearly interest
rate.

The main information you need to work out the right formula is that
with a given principal 100, the interest to be added is i*100, so that
the amount accumulated after one month has increased from 100 to
(1+i)*100.

After 1 month, the amount accumulated is 100*(1+i). If, then, a
payment is made of 100, the amount accumulated is 100*(1+i)+100.

If we repeat this process after 2 months, the amount accumulated then
would be

(1+i)*[100*(1+i)+100]+100 = 100*(1+i)^2 + 100*([1+i] + 1)

After 3 months, the amount accumulated would be

100*(1+i)^3 + 100*([1+i]^2 + [1+i] + 1)

and after 4 months,

100*(1+i)^4 + 100*([1+i]^3 + [1+i]^2 + [1+i] + 1)

= 100*(1+i)^4 + 100*([1+i]^4 - 1)/([1+i] - 1)

From this you can see (and prove) that the pattern is that after
k months, the amount accumulated is

100*(1+i)^k + 100*([1+i]^k - 1)/i

Now after time k, the amount accumulated is 1000000.  Then

1000000 = 100*(1+i)^k + 100*([1+i]^k - 1)/i)

This gives an equation relating i and k.

For a given interest rate, you solve this equation for k:

1000000*i/100 = (1+i)^(k+1) - 1,
(1+i)^(k+1) = 10000*i + 1,
(k+1)*log(1+i) = log(10000*i+1),
k = -1 + log(10000*i+1)/log(1+i).

This trick, to take the logarithms of both sides, is often useful when
the unknown appears in an exponent.

Remember that the annual interest rate is 12*i, so your 8 percent
annual rate translates to i = 2/300 = 0.0066666..., which we use in
the above formula.

Now you substitute in your interest rate and get the number of months
until the \$1,000,000.00 has accumulated.  Of course you will have to
round upwards, since the answer will probably not be an integer.  Then
the last payment will probably take you over \$1,000,000.00, as opposed
to hitting it on the nose!

For your 8 percent rate, I get 634 months, or nearly 53 years.  You
can calculate the others yourself.

If this is not clear, or not what you wanted, write back and we can
try again.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Date: 07/24/97 at 16:43:13
From: Patrick Burke
Subject: Re: Compound interest to a specific amount

Thank you. Quite helpful. Exactly what I was looking for.

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Associated Topics:
High School Interest

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