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C3 rearranging equation

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    Hi,
    Stuck on ex4c q6c:

    Show that the equation e^0.8x-1/3-2x=0 can be written in the form
    x=p ln(3-2x), stating the value of p.

    Can anyone help? thanks
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    Your question is not clear enough
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    how?
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    you need to get the exponential term on the left and the other stuff on the right.
    then think how to change into a non-exponential expression on the left.
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    (Original post by Shao Kahn)
    Hi,
    Stuck on ex4c q6c:

    Show that the equation e^0.8x-1/3-2x=0 can be written in the form
    x=p ln(3-2x), stating the value of p.

    Can anyone help? thanks
    You might like to add some brackets.
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    (Original post by steve2005)
    You might like to add some brackets.
    you're right, should be: e^-0.8x-(1/3-2x)=0
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    So its like: e^{-0.8x}-\frac13 + 2x = 0?
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    (Original post by the bear)
    you need to get the exponential term on the left and the other stuff on the right.
    then think how to change into a non-exponential expression on the left.
    right, i think i'm close

    e^-0.8x-(1/3-2x)=0

    e^-0.8x=(1/3-2x)

    ln(-0.8x)=(1/3-2x)

    ln(3-2x)=1/-0.8x

    ln(3-2x)=-1.25x

    i know i'm close because in the answers p=-1.25,
    i just can't get it into the form x=p ln(3-2x)..
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    is this clearer?.. e^-0.8x-(1/(3-2x))=0
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    (Original post by Shao Kahn)
    right, i think i'm close

    e^-0.8x-(1/3-2x)=0

    e^-0.8x=(1/3-2x)

    ln(-0.8x)=(1/3-2x)

    ln(3-2x)=1/-0.8x

    ln(3-2x)=-1.25x

    i know i'm close because in the answers p=-1.25,
    i just can't get it into the form x=p ln(3-2x)..
    taking logs is the correct way to proceed but

    ln(e-0.8x) is just -0.8x... you must also take the log of the RHS
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    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..
  12. Offline

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    (Original post by Shao Kahn)
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..
    \displaystyle e^-^{0.8x}- \frac{1}{3-2x} = 0 Finally:rolleyes:

    And you are after something in the form

    \displaystyle x = p \ln (3-2x)

    Just for you

    Is it

    \displaystyle e^-^0^.^8^x

    or

    \displaystyle e^0^.^8^x

    you are after:confused:

    Log laws

    \displaystyle a \ln (x)  = \ln (x^a)
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    (Original post by Shao Kahn)
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..



    Uploaded with ImageShack.us


    Is this it ?
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    (Original post by Shao Kahn)
    sorry, its definitely e^0.8x-(1/(3-2x))=0

    new to this..
    Right okay so it's


    (Original post by SubAtomic)
    \displaystyle e^0^.^8^x- \frac{1}{3-2x} = 0
    (P.S. SubAtomic - to do e^{0.8x} you can just do e^{0.8x} with no need to do e^0^.^8^x)


    So, Shao Kahn. As before, take the fraction to the right hand side. Natural log both sides.

    Hint: \dfrac{1}{3-2x} = (3-2x)^{-1}
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    (Original post by steve2005)



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    Is this it ?
    yes! but with e^-0.8x!
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    (Original post by Shao Kahn)
    yes! but with e^-0.8x!
    In that case, the answer is wrong (if the answer is in fact p = -1.25)
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    (Original post by Shao Kahn)
    right, i think i'm close

    e^-0.8x-(1/3-2x)=0

    e^-0.8x=(1/3-2x)

    ln(-0.8x)=(1/3-2x)

    ln(3-2x)=1/-0.8x

    ln(3-2x)=-1.25x

    i know i'm close because in the answers p=-1.25,
    i just can't get it into the form x=p ln(3-2x)..
    Here is where you went wrong.

    You should have gotten -0.8x = ln(1/3-2x) instead.
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    (Original post by hassi94)
    In that case, the answer is wrong (if the answer is in fact p = -1.25)
    I think p = 1.25




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    (Original post by steve2005)
    I think p = 1.25




    Uploaded with ImageShack.us
    Yep that's what I got - but she said the answer says -1.25 which I was saying is wrong
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    I know this post was awhile ago but:

    f(x)=e0.8x - 1/(3-2x)

    e0.8x - 1/(3-2x) = 0

    e0.8x = 1/(3-2x)

    e-0.8x = (3-2x)/1

    -0.8x = ln(3-2x)

    x= 1/(-0.8)ln(3-2x)

    therefore x=-1.25ln(3-2x)

    so P= -1.25

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