Sorry for the inconvenience, I sent the link without the invitation. It is correct now you can try it.

To get the exponent, it can be shown that for any power of two $$$h$$$ having $$$d$$$ digits there is not any other power of two having the same $$$d$$$ and the same $$$d$$$-th digit ($$$\lfloor \frac{h}{10^{d-1}} \rfloor$$$). For example, let's consider $$$256$$$ the next power of two ($$$512$$$) differs in the last digit ($$$5$$$), based on this $$$log_2(256)$$$ is the same as $$$\lceil log_2(200) \rceil$$$ because there is not any other power of two having the same number of digits and the same $$$d$$$-th digit, now let's represent this as a mathematical formula, let's call the $$$d$$$-th digit $$$r$$$, $$$\lceil log_2(r \cdot 10^{d-1}) \rceil$$$ which can be also written as $$$\lceil log_2(r) + (d-1) \cdot log_2(10) \rceil$$$ and can be easily calculated.

there is just a simple observation that for every edge you just need to consider it's most significant bit

I don't think that this is correct, what if you had $$$111_2$$$ ($$$7$$$) and $$$110_2$$$ ($$$6$$$) their product is $$$101010_2$$$ ($$$42$$$) having $$$6$$$ bits. But if you made them $$$100_2$$$ and $$$100_2$$$ ($$$4$$$) their product is going to be $$$10000_2$$$ ($$$16$$$) having only $$$5$$$ bits