# Express each power as a radical.

• Sep 5th 2012, 03:37 PM
ford2008
Express each power as a radical.
Hi guys,

1. 9^1/2
my anser:

= root of 9

2. m^-1/2

= 1
------
m^1/2

= 1
--------
root of m

3. 3k^2/5

= 5 root of 3k^2

4. ((pq^3)^4)^1/5

= 5 root of pq^12

5. x^2/3 y^1/4
-------------
x^1/2 y^1/2

= x^4/6-3/6 y^1/4-2/4

= x^1/6 y^-1/4

Is that right? If not, please explain. Thanks.
• Sep 5th 2012, 06:29 PM
topsquark
Re: Express each power as a radical.
Quote:

Originally Posted by ford2008
Hi guys,

1. 9^1/2
my anser:

= root of 9

2. m^-1/2

= 1
------
m^1/2

= 1
--------
root of m

3. 3k^2/5

= 5 root of 3k^2

4. ((pq^3)^4)^1/5

= 5 root of pq^12

5. x^2/3 y^1/4
-------------
x^1/2 y^1/2

= x^4/6-3/6 y^1/4-2/4

= x^1/6 y^-1/4

Is that right? If not, please explain. Thanks.

Your notation on 3 and 4 are possibly wrong.

3. Is this $\displaystyle 3k^{2/5}$ or $\displaystyle (3k)^{2/5}$ If it is the second one then you are correct.

4. Is this $\displaystyle \left ( \left (pq^3 \right )^4 \right )^{1/5}$ or $\displaystyle \left ( \left ( \left ( pq \right )^3 \right )^4 \right )^{1/5}$

The other ones look good.

-Dan
• Sep 6th 2012, 09:35 AM
ford2008
Re: Express each power as a radical.
Hey Dan,

No. 3. is the first one, so then I am wrong. But what is wrong then in my answer?

No. 4. is the first one. So is my answer right or not, if it is the first one?

• Sep 6th 2012, 09:54 AM
Plato
Re: Express each power as a radical.
Quote:

Originally Posted by ford2008
3. 3k^2/5
#3 $\displaystyle .~~\sqrt[5]{3k^2}$ NOT $\displaystyle 5\sqrt{3k^2}$
$4$\displaystyle .~~\sqrt[5]{p^4q^{12}}$• Sep 6th 2012, 11:27 AM earboth Re: Express each power as a radical. Quote: Originally Posted by ford2008 Hi guys, the task: Express each power as a radical. ... 3. 3k^2/5 my answer: = 5 root of 3k^2 4. ((pq^3)^4)^1/5 my answer: = 5 root of pq^12 ... Quote: Originally Posted by ford2008 ... No. 3. is the first one, so then I am wrong. But what is wrong then in my answer? No. 4. is the first one. So is my answer right or not, if it is the first one? Thanks for your help. According to your answer the 3 in your term doen't belong under the root-sign:$\displaystyle 3 \cdot k^{\frac25} = 3 \cdot \sqrt[5]{k^2}\displaystyle \left ( \left (pq^3 \right )^4 \right )^{1/5} = p^\frac45 \cdot q^{\frac{12}5}\$