IMPORTANT:
#loga^b=bloga#, as long as base of #log# stays the same
If that 6 on the left side was a 10, we could just take #log_10# of both sides (which is already on standard calculators). However, we can't use a calculator to directly compute #log_6# of anything. So we use what we can and take #log_10# of both sides.
#6^x = 3660#
#log_(10)6^x=log_(10)3660#
#xlog_(10)6=log_(10)3660#
#x=log_(10)3660/log_(10)6#
And because we've isolated #x# on one side, as well as made all #log#s base #10#, you can type in into your calculator:
#x~~4.579...#
Of course you could always put this result back into the original equation, and you should get very close to #3660# (the more decimal places you put back in, the closer you'll get)