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Posted by on Thursday, February 2, 2012 at 2:56am.

Solve the equation. Check for extraneous solutions.
ln x + ln(x-2) = 1

  • Alg 2 - , Thursday, February 2, 2012 at 7:33am

    lnx + ln(x-2) = 1 , where x>2
    ln(x(x-2)) = 1
    x^2 - 2x = e^1
    x^2 - 2x + 1 = e+1 , ( I completed the square)
    (x-1)^2 = e+1
    x - 1 = ± √(e+1)
    x = 1 ± √(e+1)
    but x > 2 , so

    x = 1 + √(e+1)

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