# Time given future value of reoccurring deposit

I've recently come across the follow formula for calculating future value with reoccurring deposits formula. However, I have been unable to solve for t. I essentially want to find how long it will take to earn a million dollars (the future value). Any ideas on this formula or how to go about finding it?

• p = initial value = 2500
• n = compounding periods per year = 12
• r = nominal interest rate, compounded n times per year = 4% = 0.04
• i = periodic interest rate = r/n = 0.04/12 = 0.00333333
• y = number of years = 5
• t = number of compounding periods = n*y = 12*5 = 60
• d = periodic deposit = 100
• What have you tried? This is basically a few steps of algebraic manipulation. Get the terms with t exponent to one side, then factor it out. Logs will let you isolate t. – JTP - Apologise to Monica Jul 4 '16 at 15:00
• @JoeTaxpayer I've tried isolating and using logs; however, I have been unable to completely isolate t. Unfortunately, the algebraic manipulation is a bit harder than expected at first glance (it seems). – BenLeim Jul 4 '16 at 15:05
• Could you share the definition of the variables in the formula? Include the definition of p, and the timing of the payments and the valuation... – DJohnM Jul 4 '16 at 16:08
• I can't post an answer as I'm away from desktop computer, if no answer by tomorrow, I'll post then. Meanwhile you can use spreadsheet formulas to solve for t, even though the inner working of that equation remains hidden . – JTP - Apologise to Monica Jul 4 '16 at 17:33
• @DJohnM I added the definitions in an edit. The total is set to 1,000,000. – BenLeim Jul 4 '16 at 17:48 Here `T` is the total (\$1 million in your case). I used natural log but of course you can use log with any base.

The answer to this question on Math SE also provides a derivation, using different variable names and missing the `i+1` factor at the end of your formula (basically meaning the first periodic deposit is delayed by one period).

Note that i + 1 = 301/300.

Let X denote (i + 1).

Then you wish to solve :

1,000,000 = 2500X^t + 100X(X^t - 1) / (1/300).

Simplifying and taking logarithms of each side gives :

ln(32.41) = (t + 1)(ln(301/300))

This give a value of about t = 1040.

This agrees with the other answer (posted by BrenBarn).