cube root (8xh^-5xs^-3)/cube root (h^-9xs^9). Please help!

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cube root (8xh^-5xs^-3)/cube root (h^-9xs^9). Please help!

Mathematics
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  • phi
do the x's mean variable x or times?
times
  • 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?

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yes!
  • phi
I would rewrite that using exponent 1/3 instead of cube root sign It means the same thing, but easier to manipulate \[\left( \frac{ 8h^{-5}s^{-3}}{h^{-9}s^{9}}\right)^\frac{1}{3} \]
  • phi
do you know how to figure out \[ \frac{h^{-5}}{h^{-9}}\]?
do you make the -9 positive by bring it up to the numerator?
  • 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.
  • 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
okay, so the question wants me to put it into kh^rs^t. I got 2h^4/3s^2
  • 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
  • phi
using the subtract rule h^-5/ h^-9 = h^(-5 - (-9)) -5 - -9 is -5+9 = +4 h^4 is the simplified version
then it'll be 4/3?
  • 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 \[ \left(2^3 h^4 s^{-12}\right)^\frac{1}{3} \]
oh is it 2h^4/3s^-4
  • 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 ) \[ 2 h^\frac{4}{3} s^{-4} \]
so (2h^4/3)/s^4
  • phi
I think no parens confused me I would write it as 2 h^(4/3) s^(-4)
  • phi
because they want it in the form k h^r s^t
oh I got the question right! thank you!
they wanted me to keep the negative exponent

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