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 November 16th, 2016, 04:24 AM #1 Newbie   Joined: Nov 2016 From: London Posts: 7 Thanks: 0 help with a question Hi all, I've been set a question and was wondering if I was doing it correctly, here it is: Maths unit 1 assignment 3 question 1 b.jpg My instinct is telling me to factorise the numbers individually but from there I'm a little lost how to simplify it further. Any help or assistance would be massively appreciated!! Cheers all, D
 November 16th, 2016, 04:44 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400 $\dfrac{(2^{1/6} \cdot 6^{1/6}) \cdot (2^{1/3} \cdot 3^{1/3})}{(2^{1/3} \cdot 6^{1/6}) \cdot 2^{1/3}}$ can you finish the simplification?
 November 20th, 2016, 09:29 AM #3 Newbie   Joined: Nov 2016 From: London Posts: 7 Thanks: 0 Hi Skeeter, thanks for the reply, Been thinking about this for a wee bit, and I've come to this: $\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$ Last edited by skipjack; November 20th, 2016 at 10:14 AM.
 November 20th, 2016, 10:15 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,951 Thanks: 1599 It can be simplified further.
November 20th, 2016, 12:38 PM   #5
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Quote:
 Originally Posted by Alavanca11 Hi Skeeter, thanks for the reply, Been thinking about this for a wee bit, and I've come to this: $\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$
If I tell you that $\displaystyle \frac{x^a}{x^b} = x^{a - b}$, what can you do with it?

-Dan

 November 21st, 2016, 02:06 AM #6 Newbie   Joined: Nov 2016 From: London Posts: 7 Thanks: 0 ooooohhhhhh!!! Of course! $\dfrac{(2^{1/6}) \cdot (3^{1/3})}{2^{1/3}}$ $=(2^{-1/6}) \cdot (3^{1/3})$ or $\dfrac{3^{1/3}}{2^{1/6}}$
November 22nd, 2016, 12:08 AM   #7
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Quote:
 Originally Posted by topsquark If I tell you that $\displaystyle \frac{x^a}{x^b} = x^{a - b}$, what can you do with it? -Dan
Would the final answer be $\dfrac{3^{1/3}}{2^{1/6}}$?

Last edited by skipjack; November 27th, 2016 at 03:39 AM.

 November 27th, 2016, 02:13 AM #8 Banned Camp   Joined: Nov 2016 From: St. Louis, Missouri Posts: 28 Thanks: 4 Math Focus: arithmetic, fractions I tried solving the equation until the 16 times 1/12 part threw me off.
 November 27th, 2016, 03:47 AM #9 Global Moderator   Joined: Dec 2006 Posts: 18,951 Thanks: 1599 It's 16 raised to the power of 1/12, not 16 times 1/12.

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