text: Okay, let's see if we can get through this.  I'll use "R" instead of "radical" just to keep it a bit shorter\n\n[log(2R2) - 3log(R3) - 3] / [log 9 - log 4 + 4]\n\nStart in the top.  Notice that 2R2 = 2(2^1/2) = 2^3/2 (2 to the 3/2 power).  Likewise, R3 = 3^1/2 (3 to the 1/2 power).  So:\n\n[log(2^3/2) - 3log(3^1/2) - 3] / [log 9 - log 4 + 4]\n\nNow remember that an exponent within a logarithm moves out to the front.  So:\n\n[(3/2)(log 2) - (3/2)(log 3) - 3] / [log 9 - log 4 + 4]\n\nNow factor out the 3/2 from each of the terms:\n\n[(3/2)(log 2 - log 3 - 2] / [log 9 - log 4 + 4]\n\nNow let's go to the denominator.  We'll rewrite the values 9 and 4 as squares, then move the exponents out like we did before.\n\n[(3/2)(log 2 - log 3 - 2)] / [log(3^2) - log(2^2) + 4]\n[(3/2)(log 2 - log 3 - 2)] / [2 log 3 - 2 log 2 + 4]\n\nNow factor 2 out of each term in the denominator.\n\n[(3/2)(log 2 - log 3 - 2)] / [2(log 3 - log 2 + 2)]\n\nOne more little thing to make this work.  We actually need the signs changed in the numerator, so let's factor out a -1 from each term (we should have factored out -3/2 earlier):\n\n[(-3/2)(log 3 - log 2 + 2)] / [2(log 3 - log 2 + 2)]\n\nNow notice that the top and bottom both have (log 3 - log 2 + 2).  Since this appears on both top and bottom, cancel it out.  This leaves us with:\n\n(-3/2) / 2\n-3/4\n\nI know this was long and drawn out--I hope you followed the steps.