Let's say that Dijkstra’s algorithm with the priority queue using a d-ary heap. if adjusting d, we can try to achieve the best runtimes for the algorithm with d being $\sim |E|/|V|$.
Then for a fixed $|V|$, what is the largest possible ratio between this runtime and the runtime of Dijkstra using a Fibonacci heap? Where knowing the Fibonacci heap: delete_min = $O(\log |V |)$, insert/decrease_key = $O(1)$ (amortized) and $|V|$ × delete_min + $(|V | + |E|) $ x insert $= O(|V|\log|V|+|E|)$.
On the other hand, d-ary heap implementation : delete_min = $O(\dfrac{d \log|V|}{\log d})$, insert/ decrease_key =$O(\dfrac{\log|V|}{\log d}$), and |V | × delete_min + (|V | + |E|) × insert = $O( (|V|·d+|E|)\dfrac{ \log|V|}{\log d})$.
As trying to follow a provide solution, but I am not sure why it reduces to $O(\dfrac{\log|V|}{\log |E|/|V|})$, in the case 1 where |E| dominates, so Dijkstra with Fibonacci heap is $O(|E|)$. How can we get the ratio as $O(\dfrac{\log|V|}{\log |E|/|V|})$ while Dijkstra with d-ary heap is $O( (|V|·d+|E|)\dfrac{ \log|V|}{\log d}$)?