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Math Help - Transform Equation

  1. #1
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    Transform Equation

    I need help to transform this for y,I am struck after so far.the formula is
    x= 3+ ln[ y ] Any help greatly appreciated
    1 - y
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jaffa View Post
    I need help to transform this for y,I am struck after so far.the formula is
    x= 3+ ln[ y ] Any help greatly appreciated
    1 - y
    i am not sure what you wrote. and it is probably why your question hasn't been answered yet. please review your posts when you make them to make sure they are understandable

    did you mean x = 3 + \ln \bigg( \frac y{1 - y} \bigg) ??
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  3. #3
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    Transform Equation

    Yes Jhevon,x= 3+ ln[y/1-y],iswhat I was trying to write, sorry I am new to
    posting on forums, but will try to get the hang of it. Hope you can help.
    Thanks.
    Last edited by Jaffa; September 28th 2008 at 03:58 AM. Reason: wrong spelling
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jaffa View Post
    Yes Jhevon,x= 3+ ln[y/1-y],iswhat I was trying to write, sorry I am new to
    posting on forums, but will try to get the hang of it. Hope you can help.
    Thanks.
    ok, i will start by giving you this rule. it is a very important rule (the definition of a logarithm)

    \log_a b = c \Longleftrightarrow a^c = b

    note that \ln x \equiv \log_e x

    can you apply that rule here to get rid of the log?
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  5. #5
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    Transform Equation

    Hi Jhevon
    Iwill show you how far I have got,

    x=3+ln[y/1-y]
    x-3=ln[y/1-y
    e^x-3=y/1-y
    e^x-3(1-y)=y
    I hope this is right so far ? Now I am stuck,please help

    Thanks
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jaffa View Post
    Hi Jhevon
    Iwill show you how far I have got,

    x=3+ln[y/1-y]
    x-3=ln[y/1-y
    e^x-3=y/1-y
    e^x-3(1-y)=y
    I hope this is right so far ? Now I am stuck,please help

    Thanks
    please use parentheses. you should have typed something like e^(x - 3) * (1 - y) = y

    but yes, you are right so far. now expand the brackets

    e^{x - 3} - ye^{x - 3} = y

    you think you can take it from here? remember, to solve for y we need to get all y's on one side of the equation. then see if we can get to a y = \cdots kind of format
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  7. #7
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    Hi Jhevon
    Still getting use to this posting,but learning !
    I don't know if Iam right ,but here goes

    e^(x-3)-ye^(x-3)=y
    e^(x-3)-y=y/e^(x-3)
    e^(x-3)=2y/e^(x-3)
    e^(x-3)^2/2=y

    Can't find away of writing inaformat like you do ?

    Thanks again for your help
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jaffa View Post
    Hi Jhevon
    Still getting use to this posting,but learning !
    I don't know if Iam right ,but here goes

    e^(x-3)-ye^(x-3)=y
    e^(x-3)-y=y/e^(x-3)
    e^(x-3)=2y/e^(x-3)
    e^(x-3)^2/2=y
    that's wrong. you cannot divide only one term on the right by e^{x - 3}.

    as i said, get all the y's together.

    \Rightarrow e^{x - 3} = y + e^{x - 3}y

    any ideas now?

    Can't find away of writing inaformat like you do ?

    Thanks again for your help
    see post #1 here
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  9. #9
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    Transform Equation

    Hi Jhevon
    I am getting there ,I think

    e^(x-3)=y+e^(x-3)y
    e^(x-3)/e(x-3)=2y
    then is ite^(x-3)-e^(x-3)/2=y

    Maybe ?
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jaffa View Post
    Hi Jhevon
    I am getting there ,I think

    e^(x-3)=y+e^(x-3)y
    e^(x-3)/e(x-3)=2y
    then is ite^(x-3)-e^(x-3)/2=y

    Maybe ?
    no

    e^{x - 3} = y + e^{x - 3}y ....................factor out y

    \Rightarrow e^{x - 3} = y(1 + e^{x - 3}) ............divide by 1 + e^x

    \Rightarrow \frac {e^{x - 3}}{1 + e^{x - 3}} = y
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