ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?

Question

anonymous · Answer

$$\sqrt[3]{s}\div s ^{2}$$

mathteacher1729 · Answer

I believed this was asked just a little while ago today...

anonymous · Answer

why is to the -1/3 and not positive 1/3?

anonymous · Answer

sorry, my bad!

anonymous · Answer

$$s ^{1/3} . s ^{-2}$$ = s$$s ^{-5/6}$$

amistre64 · Answer

$${s^{1/3} \over s^2} = {1 \over s^{2-\frac{1}{3}}}$$

amistre64 · Answer

1/s^(5/3)

anonymous · Answer

thank you, jgeorge!  Thanks so much!

anonymous · Answer

any time :)

anonymous · Answer

Amistre64, where in the heck have you been?  I have calculus problems to solve, you know!!!  ; )

amistre64 · Answer

i had to do laundry; and air up my tire; put oil in the car... you know, nonmathical stuff lol

anonymous · Answer

Well, next time do those things when I DON'T need your help!  Seriously, how have you been?  Can you help me now with the antiderivative of that exponent you gave me?  And wait a minute, why is it to the -5//6 and not -5/3?

amistre64 · Answer

he confused 3*2

amistre64 · Answer

2-(1/3) = (3*2 -1)/3

anonymous · Answer

so it is s^-5/3, then?  right?

amistre64 · Answer

yes

amistre64 · Answer

and you want to derive that?

anonymous · Answer

Need help finding the antiderivative of this if you don't mind then.  Let me try to attach an equation just to look professional, like i know what I'm doing!  ; )

amistre64 · Answer

no keeping secrets now lol

anonymous · Answer

$$1+s ^{-5/3}$$

anonymous · Answer

Need to find the antid of that!  I know it's s-something.  but it's the "something" i get stuck on, sort of!

amistre64 · Answer

$$(-5/3)s^{(-5/3 - 3/3)}$$

amistre64 · Answer

$$\frac{-5}{3 s^2 \sqrt[3]{s^2}}$$

amistre64 · Answer

$*\sqrt[3]{s}$ top and bottom to rationalise the denm

amistre64 · Answer

$$\frac{-5 \sqrt[3]{s}}{3s^3}$$

amistre64 · Answer

does that help?

anonymous · Answer

Well I am trying to figure out the first step where had -5 over all that gobbledy-gook.  I don't get that part at all!

amistre64 · Answer

its the same steps as derive $x^4$

amistre64 · Answer

$\ 'exp' * x^{(exp-1)}$

anonymous · Answer

Where did the s^2 come fronm in the denominator?  I see about the cube root, but not that.

amistre64 · Answer

lets take this one step at a time....

$$s^{(a)} \iff a*s^{a-1}$$
$$\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}$$

amistre64 · Answer

$$\frac{-5}{3s^{8/3}} \iff \frac{-5}{3s^{6/3}s^{2/3}}$$

amistre64 · Answer

$$\frac{-5}{3s^2 s^{2/3}}*\frac{s^{1/3}}{s^{1/3}}=\frac{-5s^(1/3)}{3s^2 s}$$

anonymous · Answer

Ok, I see where you are getting those numbers, but why do you have to do the 6/3 2/3 thing in the denominator in the first place?  Why do you have to break it up like that?  You can't just leave it?  Why 2 "s" 's?

amistre64 · Answer

$$\frac{-5 \sqrt[3]{s}}{3s^3}$$

amistre64 · Answer

for the same reason why we dont leave $\sqrt{36}$ as it is :)

anonymous · Answer

ok, but why the s^3?  I though it was s^2 in the denominator!  aaarrrgh!

amistre64 · Answer

$$\sqrt[3]{s.s.s.s.s.s.s.s} \iff s^2 \sqrt[3]{s^2}$$

amistre64 · Answer

$$\sqrt[3]{s^3 s^3 s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}$$
$$s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2}$$

anonymous · Answer

ok, go back up if you would to where I said,"ok, I see where you get those numbers, but why do you have to do the 6/3 2/3 thing in the denominator."  See that post of mine?  the one right before it is yours.  THATS what I don't understand!

anonymous · Answer

The s^1/3 part.

amistre64 · Answer

and the posts i just did explain that :)

anonymous · Answer

ok, then I am just an idiot, because I don't get it at all.  What to do?

amistre64 · Answer

$\frac{-5}{3s^{8/3}}$ is what we get in the denominator right?

amistre64 · Answer

your question is about the s^(8/3) and how we play with it

anonymous · Answer

Is that 3s^8/3?  It's quite small!  But yes that is my question.  How to play with it.  But i have other things I would rather play with.  Exponents AINT one of em!

amistre64 · Answer

Does $s^{8/3}$ equal $\sqrt[3]{s^8}$?

anonymous · Answer

yes, i would have to say it does.  ok, go on...  please!

amistre64 · Answer

Does $s^8$ mean: $s.s.s.s.s.s.s.s$ ?

anonymous · Answer

yes.

amistre64 · Answer

Does $s.s.s.s.s.s.s.s$ equal $(s.s.s).(s.s.s).(s.s)$ ?

Does $(s.s.s).(s.s.s).(s.s) \iff s^3 s^3 s^2$ ?

anonymous · Answer

yes, but is there a reason you split it up like that and not like (s.s.s.s).(s.s).(s.s)?

amistre64 · Answer

yes theres; our radical is a $\sqrt[3]{...}$ which means we want to group these into $s^3$ .

anonymous · Answer

OH!!!!

amistre64 · Answer

and heres why:

$\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}$

amistre64 · Answer

$$\sqrt[3]{s^3} = s$$

$$s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2 }$$

anonymous · Answer

OMGosh!  Ok, so back to the problem...I don't remember where we were even!

amistre64 · Answer

these are the basic steps that you tend to do in your head to avoid all the work of reproving them time and again lol

anonymous · Answer

Maybe in YOUR head, but not MINE!

amistre64 · Answer

yes, it does help to actually step thru them to gain insight ;)

amistre64 · Answer

ok so we are left with:

$$\frac{-5}{3s^2\sqrt[3]{s^2}}$$
and thats fine but, we maybe want to get rid of the radical in the bottom part

amistre64 · Answer

we 'know' $\sqrt[3]{s^3}=s$ so we want to multily top and bttom by $\sqrt[3]{s}$ to get rid of it in the bottom right?

anonymous · Answer

right.

amistre64 · Answer

$$\frac{-5 \sqrt[3]{s}}{3s^3}$$

anonymous · Answer

wait, if you multiply the numerator by that radical and that radical =s, why do you still have the radical sign there?

amistre64 · Answer

$$-5 * \sqrt[3]{s} = -5\sqrt[3]{s}$$

anonymous · Answer

yeah, but it's the cube root of s^3.  That just equals s I thought?

amistre64 · Answer

$$3s^2\sqrt[3]{s^2}*\sqrt[3]{s}\iff3s^2\sqrt[3]{s^2.s}\iff3s^2\sqrt[3]{s^3}\iff 3s^2.s$$
$$\implies 3s^3 $$

anonymous · Answer

Are they all this hard?  How in the heck do they expect newcomers here to calculus to know all this crap?  I really don't understand!  It's so hard!!!!!  : (

amistre64 · Answer

they expect you to have learned this in college algebra; since thats all it is

amistre64 · Answer

if they have to reteach you what you should have already learned... then it gets rather ...... misses the point of dividing mathinto teachable sections lol

anonymous · Answer

BTW your answers are always the best and easiest to work with!  As are you!  Thank you so much!  I am going to go and try towrok this problem...will you be around?  Oh, and the college algebra thing?  I got a LOW C!  I couldn't do it then; I aced calc in college 20 years ago, and now don't know a thing, but could help my daughter with high school algebra II like a pro (sort of).

amistre64 · Answer

:) youre doing fine :)

anonymous · Answer

I get the point!  Do you know of a good website with the exponent rules on it?

anonymous · Answer

Actually, somehow in calc I'm getting a 97%!

amistre64 · Answer

my sirttes are all books; but i know a good practie site for math:

interactmath.com

amistre64 · Answer

sirttes eh.... my uiniversal translator is on the fritz i think lol

anonymous · Answer

Yeah, what is a sirttes anyways?

amistre64 · Answer

it my brain cells jockeying for position lol

anonymous · Answer

Will you be around for some of today?  Can I visit you here again if I need to?

amistre64 · Answer

ill be here as long as the weather holds up; im under the veranda since the libraries are closed

anonymous · Answer

Ok, then again...thanks from the bottom of my heart for all your wonderful help!  You're the best!  Giving me all that time and patience...you're great, thanks