(x^1/6y^1/3)^-18

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(x^1/6y^1/3)^-18

Mathematics
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so I will do numerator and denominator separately
okay thank you :0]
numerator --(x^1)^-18=x^-18

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Other answers:

now what is this numerator go to?
denominator -- (6y^1/3)^-18 =(6^-18)*(y^-6)
put them together (x^-18)/(6^-18)(y^-6)
okay break this down a bit. for the numerator where did you get the x^1?
wasn't that in the question?
no its x^1/6
no its x^1/6
oh, I misread the question, sorry
its okay it was just hiding is all. haha
(x^(1/6) * y^(1/3))^-18 x^-3 y^-6
I just multiplied the exponents
theres no multiplication sign between x and y
when there is nothing, it is assumed that they are being multiplied
oh okay. so for the exponents do you just make the 3 qnd 6 equal to -18?
When you have one exponent on another like this \[(a^x)^y\] You would multiply the exponents x*y
can be rewritten as a^(x*y)
xy^1/18?
so in your case (x^(1/6) * y^(1/3))^-18
nvm i got it x^-3 and y^-6 right?
you are right
is that as simpilified as im going to get?
Yes
alrighty thank you for helping me :0]
btw do you know that negative exponent could be written as positive if you put in denominator
no i didn't but that might be important right?
yes say you have something like x^-a you can rewrite it as 1/x^a
1/ over the equation
okay yeah i saw that in a video i was recently watching okay thanks a lot!! that will help towards my test. i really really apperciate everything. :0]
also remember this x^a * y^b=z^(a+b) x^a/y^b= z^(a-b)
special cases??
no it is always the case Trying solving this x^2 * y^4
z^6?
Yes
I forgot to say this but x and y must be same
sorry my computer is running slow. in the cases you just showed me is that what your talking about?
yes
yes
2^3 * 2^2 = 8 * 4= 32 or 2^(3+2) = 2^5= 32
Will you test have log
okay for a the problem im working on now (x^1/3 /y^-2/3)^9 do i divide first?
log? like on the calculator?
since you do not know if x and y are same don't devide anything
Just do what we did on last problem
i did i got x^3/y^-6 but my book says the answer os x^3y^6
-2/3 * 9
i got -6 and i did the same for 1/3 and got 3
Both you and the book are right
1/a^-x = a^x 1/a^x = a^-x
hmmm? but i have the x^3 on top how can you fit a 1 by itself?
This is what you got x^3/y^-6 using 1/a^-x = a^x 1/a^x = a^-x x^3 * 1/y^-6 x^3 * y^6
oh okay that makes more sense.
1 * x^3 just x^3

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