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natnatwebb Group Title

Okay I'm almost there but the answer I'm getting is incorrect.

  • one year ago
  • one year ago

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  1. natnatwebb Group Title
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    \[\ln (x)+\ln (x-2)=\ln 8\]

    • one year ago
  2. natnatwebb Group Title
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    then...\[\ln (x)(x-2)=\ln 8\]

    • one year ago
  3. natnatwebb Group Title
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    then...\[\ln (x ^{2}-2x)=\ln 8\]

    • one year ago
  4. natnatwebb Group Title
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    then drop the ln... \[x^{2}-2x=8\]

    • one year ago
  5. gohangoku58 Group Title
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    going good...

    • one year ago
  6. RyanL. Group Title
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    well you e^ both sides. That's how the ln cancel. But then you are left with a quadratic equation

    • one year ago
  7. natnatwebb Group Title
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    split the two, \[x^{2}=8\] and \[-2x=8\]

    • one year ago
  8. gohangoku58 Group Title
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    thats incorrect

    • one year ago
  9. Jerry123gnzcraft Group Title
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    true ok ok

    • one year ago
  10. gohangoku58 Group Title
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    u need quadratic formula

    • one year ago
  11. Jerry123gnzcraft Group Title
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    yup

    • one year ago
  12. gohangoku58 Group Title
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    Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

    • one year ago
  13. gohangoku58 Group Title
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    x^2-2x-8=0

    • one year ago
  14. natnatwebb Group Title
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    okay gimme a second to work that out.

    • one year ago
  15. gohangoku58 Group Title
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    sure.

    • one year ago
  16. freewilly922 Group Title
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    *** and need to remember to check for valid values of x. if you are taking ln(x) and ln(x-2) what can't x be?

    • one year ago
  17. RyanL. Group Title
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    You can always complete the square. That's another way of solving this without the long work of using the quadratic

    • one year ago
  18. natnatwebb Group Title
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    2

    • one year ago
  19. natnatwebb Group Title
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    whats finishing the square? Sounds like a shortcut I was denied...

    • one year ago
  20. Jerry123gnzcraft Group Title
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    granted

    • one year ago
  21. gohangoku58 Group Title
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    shortcut only if u have some prsctice.... u wanna learn that ?

    • one year ago
  22. freewilly922 Group Title
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    ... or just factor? \[x^2-2x-8=0=(x-?)(x+?)\]

    • one year ago
  23. Jerry123gnzcraft Group Title
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    practice some more then use that form

    • one year ago
  24. gohangoku58 Group Title
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    factoring is best here..

    • one year ago
  25. gohangoku58 Group Title
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    can u factor ?

    • one year ago
  26. natnatwebb Group Title
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    I'll give it a shot! :)

    • one year ago
  27. gohangoku58 Group Title
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    sure do :)

    • one year ago
  28. Jerry123gnzcraft Group Title
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    go han!!!! go ku!!!! dominate the problem TEAMWORKKKK

    • one year ago
  29. gohangoku58 Group Title
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    ? jerry ?

    • one year ago
  30. Jerry123gnzcraft Group Title
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    GO HAN GO KU

    • one year ago
  31. RyanL. Group Title
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    \[x^2−2x−8=0=(x−A)(x+B)\\ A*B=-8\\ A+B=-2\] Find A and B

    • one year ago
  32. natnatwebb Group Title
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    Almost there....

    • one year ago
  33. RyanL. Group Title
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    You have a system of question where you can simply plug in one equation to another if you can't see it right away. Then solve for the missing variable.

    • one year ago
  34. natnatwebb Group Title
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    its (x-4)(x+2)

    • one year ago
  35. gohangoku58 Group Title
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    thats correct :)

    • one year ago
  36. gohangoku58 Group Title
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    so what are 2 values of x u got ?

    • one year ago
  37. natnatwebb Group Title
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    ten solving its 4 and -2

    • one year ago
  38. RyanL. Group Title
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    \[(x-4)(x+2)=0\]

    • one year ago
  39. natnatwebb Group Title
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    and -2 would be extraneous

    • one year ago
  40. gohangoku58 Group Title
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    but can x be -2 ?? think ....

    • one year ago
  41. gohangoku58 Group Title
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    yes, so x is only = ?

    • one year ago
  42. natnatwebb Group Title
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    4?

    • one year ago
  43. gohangoku58 Group Title
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    yup, u got it :)

    • one year ago
  44. natnatwebb Group Title
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    HAZUAHH!

    • one year ago
  45. natnatwebb Group Title
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    THanks everyone! I really appreciate it! :)

    • one year ago
  46. gohangoku58 Group Title
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    welcome ^_^

    • one year ago
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