Expert: Paul Klarreich - 11/26/2011

Question
I am wondering if there is a closed form solution, involving logarithms perhaps, to the following challenge:  

Find the smallest , positive value for N,  in which the LHS below, becomes equal to or larger than the RHS.

(((1+a)^N)-1) / (((1+b)^N) - 1) >= ( c* (a/b))

where a, b, c are positive constants, a > b, and c> 1

For example:  a=.25, b=.05, c= 1.5

then by iteration, N* = approx. 5.1

Computational support for answer:  

(((1.25)^5.1) -1) / (((1.05)^5.1)-1) > = (1.5 * (.25/.05))  

because 7.056 > 7.5

Can the general solution for N be found in any way, other than trial and error?

Answer
Questioner: Jim Monahan
Country: Massachusetts, United States
Category: Advanced Math
Private: No
Subject: is there a closed form solution
Question: I am wondering if there is a closed form solution, involving logarithms perhaps, to the following challenge:  

Find the smallest , positive value for N,  in which the LHS below, becomes equal to or larger than the RHS.

(((1+a)^N)-1) / (((1+b)^N) - 1) >= ( c* (a/b))

where a, b, c are positive constants, a > b, and c> 1

For example:  a=.25, b=.05, c= 1.5

then by iteration, N* = approx. 5.1

Computational support for answer:  

(((1.25)^5.1) -1) / (((1.05)^5.1)-1) > = (1.5 * (.25/.05))  

because 7.056 > 7.5

Can the general solution for N be found in any way, other than trial and error?
...................................................
I don't know any good way, but this might help a bit:

Simplifying a bit, your expression is:

(1+a)^N - 1
--------------- >=  ca/b
(1+b)^N - 1

Use the binomial expansion:

1 + Na + N(N-1)/2 a^2 + ... - 1
---------------------------------- >=  ca/b
1 + Nb + N(N-1)/2 b^2 + ... - 1

Na + N(N-1)/2 a^2 + ...
--------------------------- >=  ca/b
Nb + N(N-1)/2 b^2 + ...

a + (N-1)/2 a^2 + ...
--------------------------- >=  ca/b
b + (N-1)/2 b^2 + ...

That's it for now.  Sorry.