高一数学题目。请教！

Let the x - coordinate of A and B be a and b respectively. Since they are on curve y = log (8)x [ to the base 8], their y - coordinates are log(8)a and log (8)b respectively.
log (8)x = log (2)x/log (2)8 = log(2)x/log(2)(2^3) = log(2)x/3,
so A is [a, (log a)/3] and B is [b ,(log b)/3]. [Note : log is understood to be base 2 from now on.]
Since OAB is a straight line, slope of A is the same as slope of B, that is
(log a)/3a = (log b)/3b, or (log a)/a = (log b)/b........................(1)
Since C and D are vertically above A and B, their x - coordinates are the same, so x - coordinates of C and D are a and b respectively.
Since C and D are on curve y = log(2)x = log x, so the coordinates of C is (a, log a) and D is (b, log b)
Slope of OC = (log a)/a, slope of OD = (log b)/b, by result of (1), they are the same, so OCD is a straight line.
(2)
Since BC is horizontal, that means their y - coordinates are the same,
so (log b)/3 = log a, or log b = 3 log a and b = a^3
Sub. these results into (1), we get
(log a)/a = (3 log a)/a^3
1 = 3/a^2
a^2 = 3
a = sqrt 3 (- sqrt 3 is reject since a must be +ve.)
So the coordinates of A is [ sqrt 3, log(sqrt 3)/3] = [sqrt 3, (log 3)/6].