According to the conditions given in the title and the function image, let the straight line OAB equation be y=KX, the coordinates of point A be (a, Ka), and the coordinates of point B be (b, Kb). Because the straight line OAB and the function y=㏒8X intersect at point A and point B, we can bring the coordinates of point A and point B into y=㏒8X, and we can get Ka =. According to the image, the abscissas of points C and D are the same as those of points A and B, so the abscissas of points C and D are brought into the function y=㏒2X, and the ordinate of points C and D are ㏒2a and ㏒2b respectively, and the coordinates of points C and B are (a,㏒2a) and (b) respectively. Comparing the coordinates of point A and point B with those of point C and point D, we can find that 3㏒8a=㏒2a (which can be simplified by logarithmic function) and 3㏒8b=㏒2b are the same, so the coordinates of point C become (a,3㏒8a) and the coordinates of point D are (b). According to the known coordinates of c and d, the equation of straight line CD can be obtained. After simplification, a and b can be eliminated, and y=3KX can be obtained. At this point, it can be concluded that the straight line CD is the straight line passing through (0,0), that is, the origin.