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anonymous

  • 5 years ago

how do i simplify a radical expression that has a variable in it?

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  1. anonymous
    • 5 years ago
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    \[\sqrt[3]{x^7}\]

  2. amistre64
    • 5 years ago
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    make it a rational exponent :)

  3. anonymous
    • 5 years ago
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    \[x^{7/3}\]

  4. amistre64
    • 5 years ago
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    x^(7/3) = x^(6/3 + 1/3) x^2 cbrt(x)

  5. anonymous
    • 5 years ago
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    it needs to be simplified not solved

  6. amistre64
    • 5 years ago
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    that is simplified; in order to 'solve' we would have said x = some number

  7. amistre64
    • 5 years ago
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    to simplify just means to write it in another way

  8. amistre64
    • 5 years ago
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    cbrt(x^7) = x^2 cbrt(x)

  9. anonymous
    • 5 years ago
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    ok thanks but all the answers have either a 3 and an x or a 2 and a 3 and an x

  10. amistre64
    • 5 years ago
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    maybe it more accurate to write it like this: cbrt(x^7) <=> x^2 cbrt(x) :)

  11. amistre64
    • 5 years ago
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    \[\sqrt[3]{x^7} <=> x^2 \sqrt[3]{x}\]

  12. anonymous
    • 5 years ago
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    thank you so much

  13. amistre64
    • 5 years ago
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    yw :)

  14. anonymous
    • 5 years ago
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    \[\frac4{9-\sqrt6}\]

  15. anonymous
    • 5 years ago
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    rationalize the denimonator

  16. amistre64
    • 5 years ago
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    you gotta multiply by the conjugate; which is just cahngeing that - into a +

  17. amistre64
    • 5 years ago
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    4 (9+sqrt(6)) 4(9+sqrt(6)) ----------- = ----------- 81 -6 75

  18. amistre64
    • 5 years ago
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    multiply top AND bottom by the conjugate ;)

  19. anonymous
    • 5 years ago
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    thx

  20. amistre64
    • 5 years ago
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    yw :) if you post a new question in the question box, more people will get a chance to help and you wont run the risk of me not seeing it in this post :)

  21. anonymous
    • 5 years ago
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    \[\log_{5}75-\log_{5}3 \]

  22. anonymous
    • 5 years ago
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    write the expression as a single logarithm whose coefficient is 1?

  23. amistre64
    • 5 years ago
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    log(a) - log(b) = log(a/b) so, log5(75) - log(3) = log5(75/3) = log5(25)

  24. anonymous
    • 5 years ago
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    \[\sqrt{9x+22}=x \]

  25. amistre64
    • 5 years ago
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    ^2 both sides to get: 9x +22 = x^2 0 = x^2 -9x +22 0 = (x-11)(x+2) x = 11 and -2 ; but we gotta dbl check because this way can have fake results: sqrt(9(11)+22) ?= 11 sqrt(99+22) ?= 11 sqrt(121) ?= 11 11 = 11 ; that ones good -------------------------------- sqrt(9(-2) +22) = -2 .... aint no way that one works lol x = 11 is the answer

  26. anonymous
    • 5 years ago
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    u r a math god!!

  27. amistre64
    • 5 years ago
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    more of a math demigod lol

  28. anonymous
    • 5 years ago
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    lol

  29. amistre64
    • 5 years ago
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    my indian name is "runs with scissors" ....

  30. anonymous
    • 5 years ago
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    \[\log_{9}25 \]

  31. anonymous
    • 5 years ago
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    calculator keeps giving me the wrong answer

  32. amistre64
    • 5 years ago
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    change of base it.... is my guess

  33. amistre64
    • 5 years ago
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    ln(25) ----- = answer ln(9)

  34. amistre64
    • 5 years ago
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    1.464.... maybe?

  35. anonymous
    • 5 years ago
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    thanks that worked

  36. amistre64
    • 5 years ago
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    it should :)

  37. amistre64
    • 5 years ago
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    spose we have 9^x = 25 log(9^x) = log(25) x log(9) = log(25) x = log(25)/log(9) x = log9(25) its all good

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