• Jan 16th 2007, 03:00 AM
rexII143

1.) -2 4^√2 . 3^√3

2.) √10 . 2 3^√100

3.) 3^√7 . 4^√7

4.) 3^√a^2 . 7^√a^6

5.) 5 4^√a^3 . 5^√a^4

6.) 3^√x^2y . √xy^3

7.) 2 3^√16 . 6^√4

8.) 6 3^√xy^2 . √xy

9.) 5 3^√a . 4√a^3

10.) 5 3^√a^2 . 7 5^√a^6

Note:

3^√ denotes cube root, 4^√ denotes fourth root and so on.........

thank you =)
• Jan 16th 2007, 04:42 AM
AfterShock
Quote:

Originally Posted by rexII143

1.) -2 4^√2 . 3^√3

2.) √10 . 2 3^√100

3.) 3^√7 . 4^√7

Note:

3^√ denotes cube root, 4^√ denotes fourth root and so on.........

thank you =)

Hope I understand your notation correctly.

1.) -2*4^(1/4)*3^(1/3)

-2*sqrt(2)*3^(1/3)

2.) sqrt(10)*2*100^(1/3)

sqrt(10)*2*10^(2/3)

10^(1/2)*2*10^(2/3)

20*10^(1/6)

3.) 7^(1/3)*7^(1/4)

7^(7/12)

And I'll complete the others later during my lunch break- let me know if that's the notation you intended.
• Jan 16th 2007, 04:53 AM
topsquark
Quote:

Originally Posted by rexII143

1.) -2 4^√2 . 3^√3

Typically I would take this to be simplified. I'm only going to do the first one to check that I am showing you how to do what is intended. I am presuming you want this all to be put under one radical.

There are two basic concepts here:
1) $\sqrt[n]{a} = (a)^{1/n}$

2) $a^{(1/n)} b^{(1/n)} = (ab)^{(1/n)}$

(Those and getting the lcm of two numbers.)

So let's simplify things a bit and just look at the $\sqrt[4]{2} \cdot \sqrt[3]{3}$.

$\sqrt[4]{2} \cdot \sqrt[3]{3} = 2^{1/4} \cdot 3^{1/3}$

Now, take a look at the denominators of the fractional exponents: 4 and 3. we want the least common multiple of 4 and 3. This is 12. So:
$\sqrt[4]{2} \cdot \sqrt[3]{3} = 2^{1/4} \cdot 3^{1/3}$ $= 2^{3/12} \cdot 3^{4/12} = (2^3)^{1/12} \cdot (3^4)^{1/12} = (2^3 \cdot 3^4)^{1/12}$

$= (8 \cdot 81)^{1/12} = \sqrt[12]{648}$

Let me know if this is the kind of thing you were after.

-Dan
• Jan 16th 2007, 05:55 AM
rexII143
^^ yes, that's it.
• Jan 16th 2007, 06:37 AM
topsquark
Quote:

Originally Posted by rexII143
^^ yes, that's it.

Um, me (topsquark) or AfterShock? We both have valid and different ways of looking at it.

-Dan
• Jan 16th 2007, 08:42 AM
Soroban
Hello, Rex!

I'll do a few more of them . . .

Quote:

$5)\;\;5\sqrt[4]{a^3}\cdot\sqrt[5]{a^4}$

We have: . $5\cdot a^{\frac{3}{4}}\cdot a^{\frac{4}{5}} \;=\;5\cdot a^{(\frac{3}{4}+\frac{4}{5})} \;= \;\boxed{5\cdot a^{\frac{31}{20}}}$

Quote:

$6)\;\;\sqrt[3]{x^2y}\cdot\sqrt{xy^3}$

We have: . $\left(x^2y\right)^{\frac{1}{3}}\left(xy^3\right)^{ \frac{1}{2}} \;=\;(x^2)^{\frac{1}{3}}\cdot y^{\frac{1}{3}}\cdot x^{\frac{1}{2}}\cdot \left(y^3\right)^{\frac{1}{2}}$ $= \;x^{\frac{2}{3}}\cdot y^{\frac{1}{3}}\cdot x^{\frac{1}{2}}\cdot y^{\frac{3}{2}}$

. . $= \;x^{\frac{2}{3}}\cdot x^{\frac{1}{2}}\cdot y^{\frac{1}{3}}\cdot y^{\frac{3}{2}} \;=\;x^{(\frac{2}{3}+\frac{1}{2})}\cdot y^{(\frac{1}{3}+\frac{3}{2})}\;= \;\boxed{x^{\frac{7}{6}}y^{\frac{11}{6}}}$

Quote:

$7)\;\; 2\sqrt[3]{16}\cdot\sqrt[6]{4}$

We have: . $2\cdot(16)^{\frac{1}{3}}\cdot(4)^{\frac{1}{6}}\;=\ ;2\cdot(2^4)^{\frac{1}{3}}\cdot(2^2)^{\frac{1}{6}} \;=\;2\cdot2^{\frac{4}{3}}\cdot2^{\frac{1}{3}}$

. . $= \;2^{(1+\frac{4}{3}+\frac{1}{3})} \;=\;\boxed{2^{\frac{8}{3}}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I'm not sure what final form the answer should take.

For example, #6 is: . $x^{\frac{7}{6}}y^{\frac{11}{6}} \:=\:\left(x^7y^{11}\right)^{\frac{1}{6}}\:=\:\sqr t[6]{x^7y^{11}}$

And #7 is: . $2^{\frac{8}{3}}\:=\:2^{(2 + \frac{2}{3})} \:=\:2^2\cdot2^{\frac{2}{3}} \;=\;4\sqrt[3]{4}$