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Math Help - Solve for x

  1. #1
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    Solve for x

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

    Solve for x
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  2. #2
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    Quote Originally Posted by dm10 View Post
    ln(x) + ln(x-1) = 1

    Solve for x
    Use the following rules:

    1. \ln A + \ln B = \ln (AB).

    2. \ln C = D \Rightarrow e^D = C

    After applying the above rules, expand the result and arrange into a quadratic equation which you can solve for x.
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  3. #3
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    Quote Originally Posted by dm10 View Post
    ln(x) + ln(x-1) = 1

    Solve for x
    First off, we can't have x\leq 0

    Remember the rule

    \log_a(m)+\log_a(n)=\log_a(mn)?

    Using this rule we get

    \ln[x(x-1)]=1
    x(x-1)=e^1
    x(x-1)=e
    x^2-x=e
    x^2-x-e=0

    Then using the quadratic formula, namely

    If ax^2 + bx + c = 0 then x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

    we get as our solution for x...

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

    And since we know x > 0 we disregard the negative answer.

    So x = \frac{1 + \sqrt{1 + 4e}}{2}.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Use the following rules:

    1. \ln A + \ln B = \ln (AB).

    2. \ln C = D \Rightarrow e^D = C

    After applying the above rules, expand the result and arrange into a quadratic equation which you can solve for x.
    Thanks for the help.
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  5. #5
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    Quote Originally Posted by dm10 View Post
    I did use the rules but I came out with the wrong answer. The answer I got was -1/2 but the correct answer is 1/2(1 + (the square root of) 1 + 4e)
    See my response above. Can you see where you've gone wrong?
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