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Math Help - Help With Rearranging!

  1. #1
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    Help With Rearranging!

    Hey everyone!

    I have the following line which needs to be expressed for x in terms of t:

    t = ln|1-x| - ln|2-x| + c

    What I first think of doing is raising everything as a power of e. However, this causes the x's to cancel.

    eg. e^t = 1-x-(2-x) +e^c

    As you can see the x's cancel and I am stuck.

    Any help would be awesome!

    Cheers.
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  2. #2
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    Quote Originally Posted by falconed View Post
    Hey everyone!

    I have the following line which needs to be expressed for x in terms of t:

    t = ln|1-x| - ln|2-x| + c

    What I first think of doing is raising everything as a power of e. However, this causes the x's to cancel.

    eg. e^t = 1-x-(2-x) +e^c

    As you can see the x's cancel and I am stuck.

    Any help would be awesome!

    Cheers.
    t = \ln|1-x| - \ln|2-x| + c = \ln \left| \frac{1-x}{2-x} \right| + c

    \Rightarrow t - c = \ln \left| \frac{1-x}{2-x} \right|

    \Rightarrow e^{t - c} = \frac{1-x}{2-x}

    \Rightarrow A e^t = \frac{1-x}{2-x} where A = e^{-c} is just as arbitrary as c

    \Rightarrow A (2 - x) e^t = 1 - x

    \Rightarrow 2A e^t - A x e^t = 1 - x

    \Rightarrow x\left( 1 - A e^t \right) = 1 - 2A e^t

    \Rightarrow x = \frac{1 - 2A e^t}{1 - A e^t}.

    No doubt there are many careless mistakes in the above so you should check it very carefully.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    t = \ln|1-x| - \ln|2-x| + c = \ln \left| \frac{1-x}{2-x} \right| + c

    \Rightarrow t - c = \ln \left| \frac{1-x}{2-x} \right|

    \Rightarrow e^{t - c} = \frac{1-x}{2-x}

    \Rightarrow A e^t = \frac{1-x}{2-x} where A = e^{-c} is just as arbitrary as c

    \Rightarrow A (2 - x) e^t = 1 - x

    \Rightarrow 2A e^t - A x e^t = 1 - x

    \Rightarrow x\left( 1 - A e^t \right) = 1 - 2A e^t

    \Rightarrow x = \frac{1 - 2A e^t}{1 - A e^t}.

    No doubt there are many careless mistakes in the above so you should check it very carefully.

    Thanks! I completely forgot that:

    ln |a| - ln |b| = ln |a/b|

    Cheers mate.
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