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    The question is;

    Solve the following,
    ln(x+1) = 2 ln(x-1)

    I really have no idea how to go about this.. help!


    Also, it says "you may need to reject some of the values of x for which the expressions are not defined". What does this mean? I know that lnx is defined for x<0, but I don't understand how to see if this applies in a question, or how to " reject" it if it does.

    Thanks so much!
    ShengXin
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    \displaystyle a\ln f(x) = \ln f(x)^a
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    ln(\frac {a}{b}) \equiv ln(a) - ln(b)
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    Remember that a\ln b = \ln (b^a) and that if \ln p = \ln q then p = q. This will help you simplify the RHS of the equation and then get it into a quadratic that you can solve for x.

    Rejecting some of the values for which the expressions are not defined is exactly what it says on the tin (and what you said). That is, since \ln x is not defined for x &lt; 0, it must be the case that \ln (f(x)) is not defined for f(x) &lt; 0 (for some function f).

    "Rejecting" values of x just means that, if you solve your quadratic and get, say, x=k as a result, but if using x=k means that either of the logarithms is not defined, then although k is a root of the resulting quadratic it is not a root of the original equation.
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    (Original post by nuodai)
    \ln x is not defined for x &lt; 0
    or x = 0
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    first you put the squared term back

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

    then un ln it

    (x+1) = (x-1)^2

    x+1 = x^2 - 2x + 1

    x^2-3x = 0

    x(x-3)= 0

    therefore x=3, but i think that its not x=0, because if you put it back into the original equation, you end up taking a log of a negative function, which you cant do. Some one else will have to check tha.
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    So if I now have

    (x+1) = (x-1)^2

    is that right? What do I do next?
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    Sorry, I didn't see your post Banburyhammer, thank you!!
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    (Original post by ShengXin)
    Solve the following,
    ln(x+1) = 2 ln(x-1)
    ln(x+1)= ln(x-1)^2

    x+1 = (x-1)^2

    x+1 = x^2 - 2x +1

    0 = x^2 -3x

    (x - 3) (x + 0) = 0

    x - 3 = 0

    x = 3

    OR

    x + 0 = 0

    x = 0

    (as far as my understanding of the question anyway)
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    (Original post by ShengXin)
    So if I now have

    (x+1) = (x-1)^2

    is that right? What do I do next?
    Multiply out the brackets, turn in to a quadratic equation and then factorise to solve for x.
    (Quadratic being ax^2+bx+c=0
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    Note: Full solutions are discouraged in this forum. Let the OP try and post a solution only as a last resort.
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    ln0 is impossible; because of the term on the left hand side of the equation, x can't be equal to -1.

    Think about how you can get rid of the ln terms by using e... do you know how they cancel out?
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    (Original post by notnek)
    Note: Full solutions are discouraged in this forum. Let the OP try and post a solution only as a last resort.
    sorry, didn't realise, i'll remember for next time

    (good revision for me though :P )
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    (Original post by notnek)
    Note: Full solutions are discouraged in this forum. Let the OP try and post a solution only as a last resort.

    oops, didnt realise that, will do next time.
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    (Original post by ShengXin)
    So if I now have

    (x+1) = (x-1)^2

    is that right? What do I do next?
    Even my dog can solve this; expand, put equal to zero, find x.

 
 
 
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