Check whether an integer is a power of 2 without using +,- operations

Yes, that's 7 bytes. Assume for the moment that I'm using IBM codepage 907 instead of Unicode and then each character is a byte :)

0=1|2⍟⎕

i.e. 0 = mod(log(input(),2),1)

~.3/&!

Here's a solution that doesn't use the x & (x-1) method in any form. It uses x & (x/3) instead. ;-) Outputs 0 if false, 1 if true.

Explanation:

~ evals the input string to turn it into a number,
. duplicates it (for the subsequent &),
3/ divides it by three (truncating down),
& computes the bitwise AND of the divided value with the original, which will be zero if and only if the input is zero or a power of two (i.e. has at most one bit set), and
! logically negates this, mapping zero to one and all other values to zero.

Notes:

Per the clarified rules, zero is not a valid input, so this code is OK, even though it outputs 1 if the input is zero.
If the GolfScript decrement operator ( is allowed, then the 5-character solution ~.(&! posted by aditsu is enough. However, it seems to go against the spirit of the rules, if not the letter.
I came up with the x & (x/3) trick years ago on the Fun With Perl mailing list. (I'm sure I'm not the first to discover it, but I did (re)invent it independently.) Here's a link to the original post, including a proof that it actually works.

GolfScript, 11 (for 1 (true) and 0 (false))

.,{2\?}%?0>

Put the number on the stack and then run.

.,{2\?}%?0>'Yes''No'if

I love how converting 1/0 to Yes/No takes as much code as the challenge itself :D

Warning: EXTREMELY inefficient ;) It does work fine for numbers up to 10000, but once you get that high you start to notice slight lag.

Explanation: