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    Solve for x :

     log 2^{x-1} = log 3^{x+1}

    I then get :
     x-1log2=x+1log3

     \frac{x-1}{x+1}= \frac{log3}{log2}

    Dunno where to go from here though?
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    Multiply out and rearrange, the log bit just represents a number. i.e.
     x-1 = x(\frac {log 3}{log2}) + \frac {log 3}{log2}
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    Try
     (x-1)=(x+1)(\frac{log3}{log2})
    Expand and get the x terms on one side, constants on the other.

    That help at all?
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    (Original post by Sephrenia)
    Try
     (x-1)=(x+1)(\frac{log3}{log2})
    Expand and get the x terms on one side, constants on the other.

    That help at all?
    I understand that, but i really don't know where to go from there to get x.
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    (Original post by Inertia)
    I understand that, but i really don't know where to go from there to get x.
    Expand the brackets. What do you get then?
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    Think of log3 / log 2 representing a number, you could call it a, then you'd have x - 1 = ax + a, then you can get all the x's on one side
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    (Original post by benwellsday)
    Think of log3 / log 2 representing a number, you could call it a, then you'd have x - 1 = ax + a, then you can get all the x's on one side
    Yes ok, but then surely it is x-ax=1 + a

    which is x-x=1+a/a which gives 0 on the left hand side?
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    (Original post by Inertia)
    Yes ok, but then surely it is x-ax=1 + a

    which is x-x=1+a/a which gives 0 on the left hand side?
    No. You divided by a, but only divided 2 terms by a. Once you have your brackets expanded you can take x out of one side. So you will have x(****)=1+a.
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    Inertia,

    I am going to post the full solution as I think you are struggling to get to grips with this. Hope that is OK in this case.

    Spoiler:
    Show


     \log 2^{(x-1)} = \log 3^{(x+1)}

    (x-1)\log2=(x+1)\log3

    x \log 2 -\log 2=x \log 3+ \log 3

    x \log 2 -x\log 3= \log 3 + \log2

    x (\log 2 -\log 3) =\log 3 + \log 2

    x \log (\frac{2}{3})=\log 6

    x = \frac{\log 6}{\log(\frac{2}{3})}

 
 
 
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