14z^2/sqrt(z^9) = 14 sqrt(z)/z^7 , right?

Question

anonymous · Answer

@jim_thompson5910

anonymous · Answer

No

anonymous · Answer

Use the law of indices.
sqrt a = a^0.5
a^m / a^n = a^(m - n)

anonymous · Answer

I don't use indices in my class.. I don't even know what that is...

anonymous · Answer

You are rationalising? It should be 14 sqrt (z^9) / z^7

anonymous · Answer

No, finding the exact value....

jim_thompson5910 · Answer

do you know how to simplify $$\large \sqrt{z^9}$$ ??

anonymous · Answer

I was told to subtract the exponents. :P I don't know how to simplify them

jim_thompson5910 · Answer

that step will come next, but first we must simplify $$\large \sqrt{z^9}$$

jim_thompson5910 · Answer

doing that will give us this

$$\large \sqrt{z^9}$$ 

$$\large \sqrt{z^8*z}$$ 

$$\large \sqrt{z^8}*\sqrt{z}$$ 

$$\large (z^8)^{1/2}*\sqrt{z}$$ 

$$\large z^{8*1/2}*\sqrt{z}$$ 

$$\large z^{8/2}*\sqrt{z}$$ 

$$\large z^4\sqrt{z}$$

jim_thompson5910 · Answer

so 

$$\large \sqrt{z^9}$$ 

turns into 

$$\large z^4\sqrt{z}$$

hopefully you can see how

anonymous · Answer

Yes, kind of, thank you.

jim_thompson5910 · Answer

so this means that

$$\large \frac{14z^2}{\sqrt{z^9}$$

turns into 

$$\large \frac{14z^2}{z^4\sqrt{z}$$

now simplify and then rationalize the denominator