Since that formula gives "average" number of customers, it won't really tell you anything about "at least" but I think what you are saying is that you want

which is, of course, equivalent to

which is, again, equivalent to

.

Now, the simplest way to handle a complicated inequality like that is to first solve the associated

**equation**,

. Those points separate ">" from "<" and you can try a point in each interval to determine which is true for that interval.

(I see nothing to do with "logarithms" in this problem.)

How about

**you** try your best? I assume you have tried but it would help us if you would at least show

**what** you have tried!

Replacing "k" by 207000000 gives [tex]log(D)= \frac{2}{3}log(207000000)- 0.7[tex]. You should be able to calculate the right side, then take 10 to that power (I am assuming this is the common logarithm) to find D. Do the same for k= 108000000 and subtract to find "how much longer".