The basic idea: Hour 1 costs $1 and subsequent hours cost 1.07 times the previous hour.

Let H(k) be the cost of hour k. Then

**H(k) = (1.07)^(k-1)**.

Moreover, the total cost C(k) of hours 1 through k will be

C(k) = H(1) + H(2) + ... + H(k) = \sum_{j=1}^{k} H(j) = \sum_{j=1}^{k} (1.07)^(j-1) = \sum_{j=0}^{k-1} (1.07)^j.

This simplifies to:

**C(k) = (1.07^(k) - 1)*(100/7)**

Now to find the total hours achieved for a given donation amount C. It turns out that

**k = ln( 7C/100 + 1) / ln(1.07) = log_{1.07}(7C/100 + 1)**