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Compound interest to a specific amount

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


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 

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?


Patrick Burke
Training Specialist
The University of Chicago

Date: 07/24/97 at 14:20:33
From: Doctor Rob
Subject: Re: Compound interest to a specific amount


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

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 

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!   

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.

Associated Topics:
High School Interest

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