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    • Thread Starter
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    Hello,

    Could someone please help me with this problem step by step - not sure how tou get to the answer - can only get so far and :-(

    11a) Show that log[4]3 = log[2]sqrt3

    11b) Hence or otherwise solve the simultaneous equation:

    2log[2]y = log[4]3 + log[2]x

    3^y = 9^x

    Thanks for you help!
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    (Original post by christinajane)
    Hello,

    Could someone please help me with this problem step by step - not sure how tou get to the answer - can only get so far and :-(

    11a) Show that log[4]3 = log[2]sqrt3

    11b) Hence or otherwise solve the simultaneous equation:

    2log[2]y = log[4]3 + log[2]x

    3^y = 9^x

    Thanks for you help!
    How did you get on with 11a?
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    (Original post by SeanFM)
    How did you get on with 11a?
    I got, which I don't even know if its right :

    log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3

    Which seems too obvious????
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    (Original post by christinajane)
    I got, which I don't even know if its right :

    log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3

    Which seems too obvious????
    It's correct! Keep going.
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    too late ...
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    Oops forgot first line - log[4]3 = log[2]3/log[2]4 = log[2]3/2

    but dont really now why that eqals log[2]sqrt3??
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    (Original post by christinajane)
    I got, which I don't even know if its right :

    log[4]3 = 1/2log[2]3 = log[2]3^1/2 = log[2]sqrt3

    Which seems too obvious????
    There is a rule you can use to change the base of a logarithm.
    loga(b)=[logc(b)]/[logc(a)].
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    Second part I get -

    2log[2]y = log[4]3 + log[2}x 3^y = 9^x
    y = 2^x (2)
    log[2]y^2 = log[2]sqrt3 + log[2]x

    y^2 = x(sqrt3) (1)


    It follows that:

    substituting equation 2 into 1

    (2^x)^2 = xsprt3

    4x^2 = xsqrt3


    then I get stuck...
 
 
 

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