Solving recurrences using substitution method
The complexity in your proof is that you have addition in you logarithm, and you need that complexity to go away. So lets get rid of that 34, by just making $n \ge 34$, so now

$$\log\left( \frac{n + 34}{2} \right) \le \log\left( \frac{n + n}{2} \right) = \log(n).$$

I am not sure why the book chose 68, but really its the same argument. If $n \ge 68$, then $n/2 \ge 34$. Thus

$$\log\left( \frac{n + 34}{2} \right) \le \log\left( \frac{n + n/2}{2} \right) = \log\left( \frac{2n + n}{4} \right) = \log\left(\frac{3n}{4}\right) = \log(n) + \log(3/4).$$