A simple question

For x = 1, $log_{10}$ (x + 99) = $log_{10}$ 100 = 2, and it is a monotonically increasing function on the range. So by the time x reaches 2, the function is greater than 2.

We will get that $100 < x+99 < 101$ , therefore we can conclude that $x < 2 < log(x+99) < log(101)$