anonymous
  • anonymous
solve for x 1. x^5=31 2. x^4/3=-64 3. 27^2/3=x^1/2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
calculator for this one
anonymous
  • anonymous
i'll take 3.
anonymous
  • anonymous
unless you mean \[x^5=32\] in which case \[x=2\] because \[2^5=32\]

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anonymous
  • anonymous
27^(2/3) is the same as 3^2 is the same as 9 .. so 9 is left side of equation
anonymous
  • anonymous
so 9=x^(1/2) square both sides to find x x=81 ..............ans.3.
anonymous
  • anonymous
nope, it's 31 aha!
anonymous
  • anonymous
for the first one: x=31^(1/5).............ans1
anonymous
  • anonymous
and for the second one first take the cube root of -64=-4 then raise it to the power of 4 which is 256 x=256
anonymous
  • anonymous
Thanks bro :)
anonymous
  • anonymous
wait i think i did the second one wrong! however, there is something wrong with the question
anonymous
  • anonymous
well, that's what the teacher has on the paper, -__-"
anonymous
  • anonymous
you have to first take the 4th root of -64 and you can't take an even root of a negative number
anonymous
  • anonymous
unless you are dealing with complex numbers.
anonymous
  • anonymous
yes, not likely something raised to the fourth power will be negative

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