Find the position of rightmost set bit of decimal in ruby

Ruby program for Find the position of rightmost set bit of decimal. Here more information.

#  Ruby Program for
#  Find position of rightmost set bit
class BitPosition 
	#  Finding the right most active bit position
	def rightmostActiveBit(n) 
		if (n <= 0) 
			return
		end

		#    Example 
		#    ———————
		#    n    = 320
		#    n    = (00101000000)   Binary
		#    -n   = (11011000000)   (2s)
		#    Calculate log2 
		#    ——————————————
		#    Formula : log(n & -n) / log(2) + 1
		#    -----------------------------------
		#    log(320 & -320) / log(2) + 1)
		#    Here : log(320 & -320) = log(64) = 4.158883
		#           log(2)  = 0.693147    
		#    (4.158883 / 0.693147) + 1 = (7) position
		#    ————————————————————————————————————————
		#  Calculate rightmost active bits
		result = (Math.log(n & -n) / Math.log(2) + 1).to_i
		print(" Number : ", n ," Result : ", result, "\n")
	end

end

def main() 
	task = BitPosition.new()
	#  Test Cases
	#  320 = Binary(101000000)
	task.rightmostActiveBit(320)
	#  (1000) = Binary(1111101000)
	task.rightmostActiveBit(1000)
	#  (153) = Binary(10011001)
	task.rightmostActiveBit(153)
	#  (354) = Binary(101100010)
	task.rightmostActiveBit(354)
	#  160 = Binary(10100000)
	task.rightmostActiveBit(160)
end

main()

Output

 Number : 320 Result : 7
 Number : 1000 Result : 4
 Number : 153 Result : 1
 Number : 354 Result : 2
 Number : 160 Result : 6