please help me solve this. (36a^5/3)^1/2

Question

lgbasallote · Answer

$\Large \sqrt{\frac{36a^5}{3}}$

36 can be taken out of the square root since it's a perfect square..

$\Large 5\sqrt{\frac{a^5}{3}}$

a can be taken out too...

$\Large 5a^2 \sqrt{\frac{a}{3}}$

we can rationalize this which will give us

$\Large \frac{5a^2 \sqrt{3a}}{3}$

radar · Answer

$$\sqrt{36^{5/3}}$$
Is that the problem?

lgbasallote · Answer

seems the asker is not here...

radar · Answer

Sometimes the challenge is figuring out what the problem is lol.

lgbasallote · Answer

yes..that is a hard problem -__-

anonymous · Answer

thanks so the A power to 5/3 became A power to 5 over 3

radar · Answer

is the $$a ^{5/3}$$

anonymous · Answer

yes

radar · Answer

or is it $$a ^{3}\over 5$$

lgbasallote · Answer

my answer is wrong then lol..i thought it was the latter

radar · Answer

O.K.   now it is clear.

anonymous · Answer

yeah a little bit

lgbasallote · Answer

yeah..

lgbasallote · Answer

so should you or should i? lol

radar · Answer

You do good work.  Have at it, I am just an interested bystander.

anonymous · Answer

$$(36a ^{5/3})^{1/2}$$ there it is in real format

lgbasallote · Answer

the way i do this is to break it down...

$\Huge (36)^\frac{1}{2} (a^\frac{5}{3})^\frac{1}{2}$

$\Large (36)^\frac{1}{2}$  is pretty much solvable it is equal to $\large \sqrt{36} = 6$

now $\huge (a^\frac{5}{3})^\frac{1}{2}$

it is solveable by rule of exponents...$\huge (a)^{(\frac{5}{3})(\frac{1}{2})}$

that is equal to...$\huge (a)^{(\frac{5}{6})}$

combine those...

$\huge 6a^{(\frac{5}{6})}$

got it?

anonymous · Answer

thanks a lot

anonymous · Answer

got it

lgbasallote · Answer

that was a lot of latexing lol

radar · Answer

I believe it was a good answer, well presented.

lgbasallote · Answer

$\large \color{red}{t} \color{green}{h} \color{blue}{a} \color{orange}{n} \color{pink}{k} \color{yellow}{s}$

anonymous · Answer

$$^{3\sqrt{ab ^{2}}}/^{3\sqrt{4a ^{2}}b}$$ can u also help me rationalize the denominator in this one , please?

radar · Answer

Too small for these 73 year old eyes, and is that b in the denominator within the radical or is it out to the right ?

anonymous · Answer

yes sir

anonymous · Answer

within the radical

radar · Answer

$$\sqrt[3]{ab ^{2}}\over \sqrt[3]{4a ^{2}b}$$ is that it or are the 3s to the left of the radical?

anonymous · Answer

yeah, that's it

anonymous · Answer

i need ur help to rationalize the denominator...

radar · Answer

To just rationalize the denominator for this fraction, multiply both the numerator and the denominator $$\sqrt[3]{4a ^{2}b}$$
Since you have operated the same on the numerator and the denominator, you have not changed the value of the fraction, just the appearance.

anonymous · Answer

alright, thank you

radar · Answer

The final result of the fraction would appear similar to this:
$$ab \sqrt[3]{4}\over4a ^{2}b$$
which simplifies to:
$$\sqrt[3]{4}\over4a$$  I think.

radar · Answer

Good luck with these.  It is time for bed here in Missouri.

anonymous · Answer

thanks again