Work out host bits required instantly with clear inputs, formula shown and shareable results.
Host bits are the inverse of prefix length. To address H interfaces you need 2^h >= H + 2 when the network and broadcast addresses must be reserved, so h = ceil(log2(H + 2)) and the prefix is 32 - h. The spare figure shows how much of the next power of two is left over, which is the growth room the rounding gives you for free.
Host bits
h = ceil(log2(hosts + reserved)); prefix = 32 - h; usable = 2^h - reserved
510 plus the two reserved addresses is 512, which is 2^9, so nine host bits are required and the prefix is 32 - 9 = 23.
Conceptually no, but IPv6 LANs are always /64 regardless of host count, so host-bit arithmetic is only used for IPv4 and for point-to-point IPv6 links.