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clarkjets

  • one year ago

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  1. clarkjets
    • one year ago
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    |dw:1358409590944:dw|

  2. RolyPoly
    • one year ago
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    \[\log_3(x+4)+\log_3(x-2)=3\]log a + log b = log (ab) \[\log_3(x+4)(x-2)=3\] (x+4)(x-2) = 3^3

  3. clarkjets
    • one year ago
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    so would it be x^2+2x-17=0 which makes the roots x = {3.243, -5.243}?

  4. RolyPoly
    • one year ago
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    Not quite, can you check again?

  5. clarkjets
    • one year ago
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    I'm sorry I can't seem to find where im going wrong

  6. RolyPoly
    • one year ago
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    The equation you got was wrong..

  7. RolyPoly
    • one year ago
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    (x+4)(x-2) = 3^3 x^2 + 2x - 8 - 3^3 =0 ^Simplify this

  8. clarkjets
    • one year ago
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    is it not x^2+2x-17 ?

  9. RolyPoly
    • one year ago
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    what is 3^3?

  10. clarkjets
    • one year ago
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    oh my apologies I was thing of multiplying, so it is x^2+2x+19?

  11. RolyPoly
    • one year ago
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    Hmmm... x^2 + 2x - 8 - 3^3 =0

  12. clarkjets
    • one year ago
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    x^2+2x-35=0 ?

  13. RolyPoly
    • one year ago
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    Yup :) Now, solve it~

  14. clarkjets
    • one year ago
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    x=5 and x=-7 ?

  15. RolyPoly
    • one year ago
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    Yes, but you have to reject one answer there.

  16. RolyPoly
    • one year ago
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    You cannot take log for a negative number.

  17. clarkjets
    • one year ago
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    -7 correct?

  18. clarkjets
    • one year ago
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    ok thank you so much

  19. RolyPoly
    • one year ago
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    You're welcome :)

  20. RolyPoly
    • one year ago
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    And yes, you need to reject -7 :)

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