Interest theory: Fraction problem

[edcrane]

Seems like I dozed through highschool -- here's a simple problem (find n):

1 - (d^n)/n = [1 + (i^7)/7] / [1 + (i^6)/6]

The answer is n = 42 = (6*7) [where i^m and d^n are, respectively interest and discount rates compounded on interval m and n], but I don't grasp the simple logic behind it. Please advise.

[edit] oops -- I should add the following rules/equations as well:

1 + i = 1/(1 - d)

(1-[d^m]/m)*(1+[i^m]/m) = 1

(1+[i^m]/m)^m = (1+[i^n]/n)^n

(1-[d^m]/m)^m = (1-[d^n]/n)^n

 

[stapel]

Without having values for d, n, or i, I have no idea how you're supposed to get the numerical answer of "42". Sorry.

Try asking your instructor for the rest of the information.

Eliz.

 

[edcrane]

Aye -- I just realized that this was something that's supposed to be derived from the relations between d and i -- so I've edited the original post to include the relevant equations -- but I still don't quite get the logic.

 

[stapel]

If you are needing to solve this equation for some one of the variables, you still need to be given the values of all of the other variables.

If you are needing to prove some formula, then there should be no numerical information required.

Please reply with the full and exact text of the exercise, along with the complete instructions. Thank you.

Eliz.

 

[edcrane]

Geeze, I'm so retarded. I forgot to say the answer is n=42. The complete question is just the original equation posted + "find n". I assume that means that the other i's and d's somehow cancel out.

 

[stapel]

The complete question is just the original equation posted + "find n".

The original equation was:

. . . . .1 - (d^n)/n = [1 + (i^7)/7] / [1 + (i^6)/6]

I'm sorry, but without values for d and i, I see no way to "solve" this for n. Even if we had values for d and i, I'm not sure that this is solveable (algebraically), since n is both a factor and an exponent.

Eliz.

 

[edcrane]

Ok well at least I feel slightly less insane for finding this difficult. Thanks for the help -- I'll ask the TA about this monday.