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anonymous

  • 4 years ago

solve for x 1. x^5=31 2. x^4/3=-64 3. 27^2/3=x^1/2

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  1. anonymous
    • 4 years ago
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    calculator for this one

  2. anonymous
    • 4 years ago
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    i'll take 3.

  3. anonymous
    • 4 years ago
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    unless you mean \[x^5=32\] in which case \[x=2\] because \[2^5=32\]

  4. anonymous
    • 4 years ago
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    27^(2/3) is the same as 3^2 is the same as 9 .. so 9 is left side of equation

  5. anonymous
    • 4 years ago
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    so 9=x^(1/2) square both sides to find x x=81 ..............ans.3.

  6. anonymous
    • 4 years ago
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    nope, it's 31 aha!

  7. anonymous
    • 4 years ago
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    for the first one: x=31^(1/5).............ans1

  8. anonymous
    • 4 years ago
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    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

  9. anonymous
    • 4 years ago
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    Thanks bro :)

  10. anonymous
    • 4 years ago
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    wait i think i did the second one wrong! however, there is something wrong with the question

  11. anonymous
    • 4 years ago
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    well, that's what the teacher has on the paper, -__-"

  12. anonymous
    • 4 years ago
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    you have to first take the 4th root of -64 and you can't take an even root of a negative number

  13. anonymous
    • 4 years ago
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    unless you are dealing with complex numbers.

  14. anonymous
    • 4 years ago
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    yes, not likely something raised to the fourth power will be negative

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spraguer (Moderator)
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