When the ratio is $1$, why does the efficiency of the algorithm evaluate to $\mathcal{O} \left( n^d \log n \right)$?
The total work done would be:
$T(n) = \mathcal{O}(n^d) (1+1+\cdots+1^k)$
$= \mathcal{O}(n^d) (1)(\log_b n)$
where $n$ is the size of the problem, $d$ the efficiency exponent, $k$ the height of the tree and $b$ the factor the size of the problem is reduced by at each iteration
Therefore wouldn't the efficiency be $\mathcal{O}(n^d) \log_b n$ as opposed to $\mathcal{O}(n^d \log n)$?