I solved the above recurrence using master theorem and applied case $2$ to solve it.
However in the final answer I have $T(n) = \Theta(\log^{(k+1)} n)$ . what should happen to $k+1$? because the final answer is $T(n) = \Theta(\log n)$
If someone has a different approach, please do share.
Best Answer
Here we have $$T(n) = aT\left(\frac nb\right) + n^c $$ where $a=1$, $b=\frac32$, and $c=0$. Then $$\log_b a = \log_{\frac32} 1 = 0 = c,$$ so $$ T(n)\in\Theta(\log n).$$ In particular, using your reference, $k=0$ works, as $n^0\log^0 n = 1$ and $$1\in\Theta(n^0\log^0n). $$