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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 November 22nd, 2014, 01:52 AM #1 Newbie   Joined: Nov 2014 From: somewhere Posts: 3 Thanks: 0 Simplification Hi, can someone simplify this for me please. I seem to miss something when doing it $\displaystyle \frac{(3a^2b^3)^{1/2} * (4a^2b^6)^{-1/3}}{(24a^2b^2)^{1/2}}$ The answer is: $\displaystyle \frac{1}{2^{13/6}a^{2/3}b^{3/2}}$ I think that I'm doing something wrong with the 24 or something like that Thank you for your time. Last edited by skipjack; November 22nd, 2014 at 03:54 AM.
 November 22nd, 2014, 04:00 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 The answer you posted is correct. If you post your working, we can identify any mistakes you made. You need to cancel 3 at some stage, so the 24 becomes 8 (or 2³).
 November 22nd, 2014, 04:33 AM #3 Newbie   Joined: Nov 2014 From: somewhere Posts: 3 Thanks: 0 Thanks for the reply I'll post the first steps and you can check if it's good or not and i'll move on to the rest because my final answer is completely different from the correct one. $\displaystyle = \frac {3^\tfrac{1}{2} ab^\tfrac{3}{2} * 4^\tfrac{-1}{3}b^{-2}}{(2^3)^\tfrac{1}{2}ab} =\frac {3^\tfrac{1}{2} ab^\tfrac{3}{2} }{(2^3)^\tfrac{1}{2}ab * 4^\tfrac{1}{3}b^{2}}$
 November 22nd, 2014, 08:30 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 Why did you replace 24 with $2^3$? Why did $a^2$ in the second part of the numerator just disappear instead of becoming $a^{-2/3}$?
 November 22nd, 2014, 11:14 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 $\displaystyle 24^{1/2} = 3^{1/2}8^{1/2}$

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