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    how do i solve this:
    ln2x - x = 0
    if i factor x out, i get x(ln2 -1) 0
    so i'm confused on how to solve the ln bit to get e^1/2??

    EDIT : i worked out i made a mistake when factorising, it should have been x(lnx2 - 1) instead. For curiosity's sake, why isn't x is removed from lnx as well when factorising?
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    (Original post by elvss567)
    how do i solve this:
    ln2x - x = 0
    if i factor x out, i get x(ln2 -1) 0
    so i'm confused on how to solve the ln bit to get e^1/2??
    Didn’t you ask the exact same question before and gave me rep for my answer? So it made sense...?
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    [QUOTE=RDKGames;74465352]Didn’t you ask the exact same question before and gave me rep for my answer? So it made sense...?[/QUOTE

    i edited the question if you look^^
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    (Original post by elvss567)
    how do i solve this:
    ln2x - x = 0
    if i factor x out, i get x(ln2 -1) 0
    so i'm confused on how to solve the ln bit to get e^1/2??

    EDIT : i worked out i made a mistake when factorising, it should have been x(lnx2 - 1) instead. For curiosity's sake, why isn't x is removed from lnx as well when factorising?
    Your factorisations of:

     \ln(2x) - x,

    are not correct and you can not do this as ln(x) is an operator on x.

    It is like doing sin(3x) implying x( sin(3) ), which IS NOT generally correct.

    ----------------------------------------------------------------------------------------------------

    As for your equation, it depends on interpretation, as we could have:

     \ln \lvert 2x \rvert - x = 0, (note the modulus signs),

    for which there is a solution but I do not think x can be found algebraically (anyone else can help here?).

    I will give a tip that the exponential functions and ln are inverses of each other.
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    (Original post by simon0)
    Your factorisations of:

     \ln(2x) - x,

    are not correct and you can not do this as ln(x) is an operator on x.

    It is like doing sin(3x) implying x( sin(3) ), which IS NOT generally correct.

    ----------------------------------------------------------------------------------------------------

    As for your equation, note there are no real solutions (try graphing the function) but I will give a tip that the exponential functions and ln are inverses of each other.
    got it! Thankyou
 
 
 
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