Question

How to solve for this: (3√7)/(√3×∜2^3×5)

Answer 1

\[\sf \frac{3\sqrt{7}}{\sqrt{3}\cdot\sqrt[4]{2^3} \cdot 5}\]\[\sf 3 = \sqrt{3} \cdot \sqrt{3}\] so...

Answer 2

\[\sf \frac{\sqrt{3} \cdot \sqrt{7}}{\sqrt[4]{2^3}\cdot 5} = \frac{\sqrt{21}}{\sqrt[4]{2^3} \cdot 5} \]

Answer 3

That's pretty much all you can simplify since there is no common base between the 2 and 5 you can combine those two with.

Answer 4

is it possible to rationalize the denominator?

Answer 5

You understand how I got this, yes?

Answer 6

I don't think so, no.

Answer 7

if it was something like \(\sf \sqrt{2} \cdot \sqrt{7}\) that would be understandable.

Then you would get \(\sqrt{14}\)

See how that works?

Answer 8

The 2 is to a different root, the 4th root, whereas 5 is just...well, 5.

Answer 9

Ohhh okay I got it. Thank you!
What if I needed to simplify the fourth root of 6 exponent 34? How would I go about doing that?

Answer 10

Hmm.... \(\sqrt[4]{6^{34}}\) ?

Answer 11

Yes

Answer 12

I'll let @ganeshie8 explain this better, I'm solving this a bit on whim :\

Answer 13

although, this might help http://www.mathsisfun.com/numbers/nth-root.html

Answer 14

Ohh, I kind of get it.\[\sf \sqrt[4]{6^{34}}= 6^{34/4}\]

Answer 15

The basic format for these "nth" roots is \(\sqrt[m]{x^n} = x^{n/m}\)

Answer 16

Where x is any number, like 6 or 13, 14, etc..

Answer 17

So : \(6^{34/4} = 6^{17/2} = 6^{8+1/2}\)

You could distribute the power to the base, so you'd get \(6^8 \cdot 6^{1/2}\) or more simply \(6^8 \sqrt{6}\)

Answer 18

And remember, \(a^m \cdot a^n = a^{m+n}\) , That's how I was able to split the two powers.

Answer 19

@Jhannybean I understand now. Thank you so much!

Answer 20

Yay :)