Let’s imagine that the number of rabbits in a field doubles each month. If we start with 6 rabbits, how many rabbits would be in the field after 10 weeks, given that 4 weeks = 1 month?
I would think the answer would be $$6 * 2^{2.5}$$ which equals a non-integer answer, but a colleague of mine argues that the answer would be $$(6 * 2^2)+ (6 * 2^2 * \frac{1}{2})$$ which is an integer answer but I think is incorrect. Can someone explain the correct way of thinking about this problem and which answer is correct, or if even neither of the solutions are correct?
I'm not asking how to calculate non-integer exponents, but rather when their use is appropriate.
1)Logically, the number of rabbits at any given time should be an integer.
2)You concluded that the function that outputs the number of rabbits as function of time(where the units here are in month) is $$f(t)= 6*2^t $$ This is a false conclusion since number of rabbits cannot be non-integer for any $t$.
3) This problem does not have a unique solution, since the constraints of the problem are not strong enough. For example consider
$$f(t) = \begin{cases} 6*2^t, & t\in\Bbb N\ \\ c, & c\in\Bbb N\ \end{cases}$$
Such a function satisfies the problem. However, one can further impose that $f(t)$ is non-decreasing, so that at any instant in time, the number of rabbits is equal to or greater than the number of rabbits in the past instants, that is $$f(t+\epsilon) \ge f(t)$$
Again, this would not still lead to a unique solution. But one such solution is $$f(t) = \begin{cases} 6*2^t, & t\in\Bbb N\ \\ \dfrac{6*(2^{m+1}+2^{m})}{2}, & m<t<m+1, \quad m\in\Bbb N\ \end{cases}$$
I could come up with other solutions.
So the information given about the problem is not complete.