Proof of the master thorem case with floors and b = 2
In fact, let $b=1.5$, we will have $(1-\frac{1}{b-1})&lt;0$, in which case $(1-\frac{1}{b-1})^{\log_ba}$ does not even make much sense!

The problem can be traced to when the following too-relaxing relaxation was made. 

$$1-\frac{b^{j}}{n}\frac{b}{b-1} \geq 1-\frac{b^{\lfloor \log_bn \rfloor -1  }}{n}\frac{b}{b-1}$$



Can we find $\Theta(\log n)$ times when the left side is greater than, for example, $\frac12$? That should be enough to generate the factor $\log n$ needed. 

Let us solve the following equation for the index $j$.
$$1-\frac{b^{j}}{n}\frac{b}{b-1} \gt \frac12$$
We get 
$$j \lt \log_b\left(\frac{n(b-1)}{2b}\right)=\log_bn-\log_b\frac{2b}{b-1}=\frac{\log n}{\log b}-c_0$$
where constant $c_0=\log_b\frac{2b}{b-1}$.

The above should be enough to set you moving for a complete proof, although there might be a couple of other minor technicalities ahead.