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BlingBlong

  • 3 years ago

How do I sole ln(x^(2) - x - 2) = 2 + ln(x+1) This is as far as I have gotten ln((x^(2) - x - 2)/(x+1)) = 2 (x^(2) - x - 2)/(x+1) = e^(2) then I divided because I was too lazy to do algebra (x+1)(x-2) = e^(2) = x^(2)-x-2-e^(2) = 0 I'm not allowed to use a calculator I was thinking I could just sub these numbers into the quadratic formula to come up with solutions would that be the right course of action?

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  1. satellite73
    • 3 years ago
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    \[x^2-x-2=(x+1)(x-2)\] so you can cancel either first or last

  2. BlingBlong
    • 3 years ago
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    The script displaying Latex type is crashing blah

  3. satellite73
    • 3 years ago
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    probably easier to do it first, \[\ln((x+1)(x-2))=\ln(x+1)+\ln(x+2)=2+\ln(x+2)+\] \[\ln(x-1)=2\] \[x-2=e^2\] etc

  4. satellite73
    • 3 years ago
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    if you can't see the latex i say it in english the x + 1 cancels, you don't get a quadratic

  5. BlingBlong
    • 3 years ago
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    I will brb going to restart my browser so I can read what you wrote

  6. PROSS
    • 3 years ago
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    |dw:1326938853539:dw|

  7. BlingBlong
    • 3 years ago
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    wow I can't believe I didn't see that lol, I need to get some sleep thanks :D

  8. satellite73
    • 3 years ago
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    that is the ugliest looking 2 i have ever seen!

  9. PROSS
    • 3 years ago
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    lol

  10. BlingBlong
    • 3 years ago
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    next time I will factor first

  11. satellite73
    • 3 years ago
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    @pross, i think you turned to 2 into an 8

  12. PROSS
    • 3 years ago
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    heh, my mouse didn't let go....

  13. satellite73
    • 3 years ago
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    yeah i made a typo too, and put \[x-1\] when it should have been \[x-2\] but the method is good

  14. PROSS
    • 3 years ago
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    ah now i see what you are talking about...first 2 looked like and 8 and i kept it in the next line.....oops...

  15. PROSS
    • 3 years ago
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    you are right...should be x=2+e^2

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