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Math Help - Logarithmic Equations Problem

  1. #1
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    Logarithmic Equations Problem

    ln X + ln X^2 = 3


    Help I'm lost again =/


    Also

    Log X + Log(x - 21) = 2
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by JonathanEyoon View Post
    ln X + ln X^2 = 3


    Help I'm lost again =/
    I'll assume you mean:

    <br />
\ln(x) + \ln(x^2) =3,

    Well:


    <br />
\ln(x) + \ln(x^2) = \ln(x \times x^2)=\ln(x^3)=3\ln(x),

    so \ln(x)=1, and so x=e.

    RonL
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  3. #3
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    Unhappy

    Quote Originally Posted by CaptainBlack View Post
    I'll assume you mean:

    <br />
\ln(x) + \ln(x^2) =3,

    Well:


    <br />
\ln(x) + \ln(x^2) = \ln(x \times x^2)=\ln(x^3)=3\ln(x),

    so \ln(x)=1, and so x=e.

    RonL


    Uhhmmm... could you explain it a little easier
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by JonathanEyoon View Post
    Uhhmmm... could you explain it a little easier
    The logarithm property that was the original reason the were invented:

    <br />
\ln(A)+\ln(B)=\ln(A\times B)<br />

    so:

    <br />
\ln(x) + \ln(x^2)=\ln(x \times x^2)=\ln(x^3)<br />

    The second logarithm property needed is:

    <br />
\ln(x^n) = n\ln(x)<br />

    (which is realy just the previous property in disguise),

    so:

    <br />
\ln(x) + \ln(x^2)=\ln(x \times x^2)=\ln(x^3)=3\ln(x)<br />

    Therefore we have:

    3 \ln(x)=3

    and so:

    \ln(x)=1.

    But the number who's log (to any base) is 1 is the base itself, and since we have logs to the base e here; x=e

    RonL
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  5. #5
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    Quote Originally Posted by JonathanEyoon View Post
    \log x + \log(x - 21) = 2
    Do the same that CaptainBlack's previous post.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    Do the same that CaptainBlack's previous post.



    The equation should come out to x^2 - 21x -2 right?
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  7. #7
    Member Jonboy's Avatar
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    You are almost correct Jonathan (hey that's my name too!).

    So we have: logx\,+\,log(x\,-\,21)\,=\,2

    Which goes to: log(x(x\,-\,21))\,=\,2

    Now raise both to a base of 10: x^2\,-\,21x\,=\,100

    You can solve the quadratic, make sure to check that both solutions work.
    Last edited by Jonboy; July 24th 2007 at 01:37 PM. Reason: went to quick
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  8. #8
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    Quote Originally Posted by Jonboy View Post
    You are almost correct Jonathan (hey that's my name too!).

    So we have: logx\,+\,log(x\,-\,21)\,=\,2

    Which goes to: log(x(x\,-\,21))\,=\,2

    Now raise both to a base of 10: x^2\,-\,21x\,=\,100

    You can solve the quadratic, make sure to check that both solutions work.

    Thanks so much I totally forgot that Log has a natural base of 10. Appreciate it!
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  9. #9
    Member Jonboy's Avatar
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    No prob. Make sure that when you do problem like these you do them step by step and don't rush like me... it'll cause a slight but critical error. Luckily I caught myself. Good luck!
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  10. #10
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    Quote Originally Posted by Jonboy View Post
    No prob. Make sure that when you do problem like these you do them step by step and don't rush like me... it'll cause a slight but critical error. Luckily I caught myself. Good luck!


    I need help with one more problem =/

    ln(4x - 2) - ln4 = -ln(x - 2)

    I'm having trouble setting up so i can solve =/


    Nevermind I solved it
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  11. #11
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    Quote Originally Posted by JonathanEyoon View Post
    I need help with one more problem =/

    ln(4x - 2) - ln4 = -ln(x - 2)

    I'm having trouble setting up so i can solve =/


    Nevermind I solved it
    In case anyone else needs the help on it:
    ln(4x - 2) - ln4 = -ln(x - 2)

    ln \left ( \frac{4x - 2}{4} \right ) = ln[(x - 2)^{-1}]

    ln \left ( \frac{4x - 2}{4} \right ) = ln \left ( \frac{1}{x - 2} \right ) <-- Take both sides to the e power

    \frac{4x - 2}{4} = \frac{1}{x - 2}

    (4x - 2)(x - 2) = 4

    4x^2 - 10x + 4 = 4

    4x^2 - 10x = 0

    2x(2x - 5) = 0

    So either
    2x = 0 ==> x = 0
    or
    2x - 5 = 0 ==> x = \frac{5}{2}

    Substitution into the original equation show that both solutions work.

    -Dan
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