Thank you Hugo.

Sergey gaved me the same answer - it is what i was looking for. i need this to change the solution in my other thread about BitAnd(). i'll let you know what i come up with when i am done.

thanks again for your help.


>If I remember correctly this is what you are looking for:
>
>log(1024)/log(2)
>
>That is the n-based log of a number is the e based log of the number divided by the e-based log of n (sorry about the wording, if a mathematician reads this I am sure (S)He will feel the urge to kill me!)
>
>LOG_n(x) = LOG(x) / LOG(n)