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natnatwebb

  • 2 years ago

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

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  1. natnatwebb
    • 2 years ago
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    \[\ln (x)+\ln (x-2)=\ln 8\]

  2. natnatwebb
    • 2 years ago
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    then...\[\ln (x)(x-2)=\ln 8\]

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

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

  5. gohangoku58
    • 2 years ago
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    going good...

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

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

  8. gohangoku58
    • 2 years ago
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    thats incorrect

  9. Jerry123gnzcraft
    • 2 years ago
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    true ok ok

  10. gohangoku58
    • 2 years ago
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    u need quadratic formula

  11. Jerry123gnzcraft
    • 2 years ago
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    yup

  12. gohangoku58
    • 2 years ago
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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: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

  13. gohangoku58
    • 2 years ago
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    x^2-2x-8=0

  14. natnatwebb
    • 2 years ago
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    okay gimme a second to work that out.

  15. gohangoku58
    • 2 years ago
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    sure.

  16. freewilly922
    • 2 years ago
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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?

  17. RyanL.
    • 2 years ago
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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

  18. natnatwebb
    • 2 years ago
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    2

  19. natnatwebb
    • 2 years ago
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    whats finishing the square? Sounds like a shortcut I was denied...

  20. Jerry123gnzcraft
    • 2 years ago
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    granted

  21. gohangoku58
    • 2 years ago
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    shortcut only if u have some prsctice.... u wanna learn that ?

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

  23. Jerry123gnzcraft
    • 2 years ago
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    practice some more then use that form

  24. gohangoku58
    • 2 years ago
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    factoring is best here..

  25. gohangoku58
    • 2 years ago
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    can u factor ?

  26. natnatwebb
    • 2 years ago
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    I'll give it a shot! :)

  27. gohangoku58
    • 2 years ago
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    sure do :)

  28. Jerry123gnzcraft
    • 2 years ago
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    go han!!!! go ku!!!! dominate the problem TEAMWORKKKK

  29. gohangoku58
    • 2 years ago
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    ? jerry ?

  30. Jerry123gnzcraft
    • 2 years ago
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    GO HAN GO KU

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

  32. natnatwebb
    • 2 years ago
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    Almost there....

  33. RyanL.
    • 2 years ago
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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.

  34. natnatwebb
    • 2 years ago
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    its (x-4)(x+2)

  35. gohangoku58
    • 2 years ago
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    thats correct :)

  36. gohangoku58
    • 2 years ago
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    so what are 2 values of x u got ?

  37. natnatwebb
    • 2 years ago
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    ten solving its 4 and -2

  38. RyanL.
    • 2 years ago
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    \[(x-4)(x+2)=0\]

  39. natnatwebb
    • 2 years ago
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    and -2 would be extraneous

  40. gohangoku58
    • 2 years ago
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    but can x be -2 ?? think ....

  41. gohangoku58
    • 2 years ago
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    yes, so x is only = ?

  42. natnatwebb
    • 2 years ago
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    4?

  43. gohangoku58
    • 2 years ago
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    yup, u got it :)

  44. natnatwebb
    • 2 years ago
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    HAZUAHH!

  45. natnatwebb
    • 2 years ago
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    THanks everyone! I really appreciate it! :)

  46. gohangoku58
    • 2 years ago
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    welcome ^_^

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