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Math Help - help understand answer

  1. #1
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    help understand answer

    solve ln(x+2)+ln(x-1)=1

    it can be written as (x+2)(x-1)=e apparently
    What rule is this?

    and the answer is x = (sqrt[9+4e]-1)/2
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  2. #2
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    note the sum rule for logs ...

    \log(a) + \log(b) = \log(a \cdot b)


    \ln(x+1) + \ln(x-1) = 1

    \ln[(x+1)(x-1)] = 1

    change the log equation to an exponential ...

    e^1 = (x+1)(x-1)
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  3. #3
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    o ok. how do u change it to an exponential? just put e as the base and it gets rid of ln?

    and how did he get that answer? i put it into the quadratic formula but am not getting that same one.
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  4. #4
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    Quote Originally Posted by Shadowfox1 View Post
    o ok. how do u change it to an exponential? just put e as the base and it gets rid of ln?

    and how did he get that answer? i put it into the quadratic formula but am not getting that same one.
    You should know that \log_A B = C is equivalent to  A^C = B.
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  5. #5
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    (x+2)(x-1) = e

    x^2 + x - 2 - e = 0

    x^2 + x - (2+e) = 0

    a = 1, b = 1, c = -(2+e)

    x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot -(2+e)}}{2}

    x = \frac{-1 \pm \sqrt{9+4e}}{2}

    since x > 1 ...

    x = \frac{-1 + \sqrt{9+4e}}{2}
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  6. #6
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    ooooooo very true. thx a lot
    Last edited by Shadowfox1; October 14th 2008 at 03:37 PM.
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  7. #7
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    wait, but how do u know x>1?
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  8. #8
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    \ln(x+2) + \ln(x-1) ... what is the domain?
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  9. #9
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    o ok, i see. I can never find the domain without a calculator.
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  10. #10
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    I can never find the domain without a calculator.
    learn the concept of domain of a function ... the "machine" will put you at a great disadvantage.
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  11. #11
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    yea true, i should. its just hard when you get stuff involving e and log and stuff like that
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