K, so for questions like "you need 8 subnets on a class C" what do you do?

I thought it was 2^[bits needed to make 8] so for 8 subnets you need 4 bits to make 8(128 64 32 16 8 4 2 1), so 4 bits for network, 4 bits for host... so you have 8 subnets with 14 available hosts but that's wrong innit, cos the right subnet would give 8 subnets with 30 usable hosts?

I'll try explaining in a simpler way - this is going from how a book explained it -

192.168.1.0, need 8 networks
8 in binary is 00001000, so 4 bits are needed
Subnet is 255.255.255.1111000, or 255.255.255.240
Usuable IPs size is 254-240 = 14
Total subnets = 2^4 = 16
But 2^3 = 8 and that's what we want?