2,7,8 puzzle and 6 N puzzle

So here is a puzzle I herd a while back and I started thinking about it in class earlier today. The puzzle is "using the numbers 2, 7, and (0-9) make an equation that equals 8. You may use +,-,*,/,decimals, (), and powers." For example, for equation 3 I could have 7+3-2=8. Post what equations you get for 0 through 9. Extra credit for 11 through 29, additionally you may use any mathematical function you know of (I know there is a solution for 0-9, but I don't know/haven't figured out some for everything from 11-29).


So here is a second puzzle I just herd from my friends, I am not sure it is fully do-able, and I couldn't find the puzzle on line to confirm. So using exactly 6 N's per equation (where N is any integer besides 0) you must make an equation which equals 0-9. For example the solution to one could be N*N*N/(N*N*N)=1. You may use any mathematical functions you wish, however any that involve a number must have an N as the number and that N counts toward your 6. (for example Log would need to be Log base N and would leave you 5 more N's. Also integrals and derivatives require a dN and this N counts toward your 6). My friends and I could only get 0-7 so we think there may be a rule we are missing but I figured I would post this too and hopefully find out what I am missing.

Good luck

>>7
6=((N+N+N)/N)! * N/N
7=((N+N+N)/N)! + N/N

As I said, I am only stuck on 8 and 9 however at this point I am assuming they cannot be done unless I am given another special function (free derivatives or 10^X or something)