1. phi

do the x's mean variable x or times?

2. anonymous

times

3. phi

ok. people don't use x in algebra to mean times. too confusing. (8xh^-5xs^-3)/cube root (h^-9xs^9) $\frac{ \sqrt[3]{8h^{-5}s^{-3}}}{ \sqrt[3]{h^{-9}s^{9}}}$ is this it?

4. anonymous

yes!

5. phi

I would rewrite that using exponent 1/3 instead of cube root sign It means the same thing, but easier to manipulate $\left( \frac{ 8h^{-5}s^{-3}}{h^{-9}s^{9}}\right)^\frac{1}{3}$

6. phi

do you know how to figure out $\frac{h^{-5}}{h^{-9}}$?

7. anonymous

do you make the -9 positive by bring it up to the numerator?

8. phi

you could. 1/h^-9 means the same as h^9 there are "short cut rules" when multiplying or dividing the same base with different exponents.

9. phi

the simplest rule is an extension of this idea: $h \cdot h = h^2$ if we put in the exponents to make it more obvious $h^1 \cdot h^1 = h^{1+1}= h^2$ we add the exponents. another example: h*h*h * h*h using the "short version" that is obviously h^5 it is also h^3 * h^2 and we see adding 3+2 gives us the right answer: h^5

10. anonymous

okay, so the question wants me to put it into kh^rs^t. I got 2h^4/3s^2

11. phi

on the other hand $\frac{h\cdot h}{h} = h$ and that is $\frac{h^2}{h^1} = h^{2-1}= h^1= h$ we subtract top exponent minus bottom exponent

12. phi

using the subtract rule h^-5/ h^-9 = h^(-5 - (-9)) -5 - -9 is -5+9 = +4 h^4 is the simplified version

13. anonymous

then it'll be 4/3?

14. phi

now do the s's s^-3 / s^9 the new exponent will be -3 - 9 = -12 so s^-12 and though not obvious, 8= 2^3 we have so far $\left(2^3 h^4 s^{-12}\right)^\frac{1}{3}$

15. anonymous

oh is it 2h^4/3s^-4

16. phi

and we can take the cube root of each term inside notice the s^-4 is up top. (if we put it down below it will become 1/s^4 ) $2 h^\frac{4}{3} s^{-4}$

17. anonymous

so (2h^4/3)/s^4

18. phi

I think no parens confused me I would write it as 2 h^(4/3) s^(-4)

19. phi

because they want it in the form k h^r s^t

20. anonymous

oh I got the question right! thank you!

21. anonymous

they wanted me to keep the negative exponent