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    Anyone who doesn't have one should really buy one, they are only about 10 pounds each on amazon. Although be careful, certain calculators aren't allowed and you should really check the models on edexcel.
    look at page 14 of this document for full details
    http://www.jcq.org.uk/attachments/pu...CE%2009-10.pdf
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    As others have said, just a little mistake, all the working is correct.
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    Guess what - 2 more problems I can't solve

    Find the values of x for which:

     log_3x + 2log_x3 = 3

    I've converted them both into base 10 (I think)

     \dfrac{logx}{log3} + 2\dfrac {log3}{logx} = 3

    :confused:
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    (Original post by Scottishsiv)
    Anyone who doesn't have one should really buy one, they are only about 10 pounds each on amazon. Although be careful, certain calculators aren't allowed and you should really check the models on edexcel.
    look at page 14 of this document for full details
    http://www.jcq.org.uk/attachments/pu...CE%2009-10.pdf
    I have fx-83 - how do I do this?!
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    (Original post by Wilko94)
    Guess what - 2 more problems I can't solve

    Find the values of x for which:

     log_3x + 2log_x3 = 3

    I've converted them both into base 10 (I think)

     \dfrac{logx}{log3} + 2\dfrac {log3}{logx} = 3

    :confused:

    dont forget your log rules...

     log_3x + 2log_x3 = 3

     log_3x + log_x(3^2) = 3

      \dfrac{logx}{log3} +\dfrac {log9}{logx} = 3

    which would be ( i think)

     log\dfrac {x}{3} + log\dfrac {9}{x} = 3

    Could be mistaken, though?
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    does anyone know how to do this equation
    2logx=log4 + log (2x+5)
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    (Original post by Wilko94)
    I have fx-83 - how do I do this?!
    It is the button underneath the on button, one that says log. The first box you input the base and in the second you input the output.

    so for example

    3^2 = 9

    So all you do is you input log to base three with an output of 9 and then you should get two as the answer. Just to make sure, does your calculator look like the one in the link shown?

    http://www.google.co.uk/imgres?imgur...:0&tx=59&ty=29
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    (Original post by princess271)
    does anyone know how to do this equation
    2logx=log4 + log (2x+5)
    Have you done anything on it?

    these may help:

     alog x = log x^a

log a + log b = log ab

log a - log b = log \dfrac{a}{b}


    _Kar.
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    (Original post by Wilko94)
    Guess what - 2 more problems I can't solve

    Find the values of x for which:

     log_3x + 2log_x3 = 3

    I've converted them both into base 10 (I think)

     \dfrac{logx}{log3} + 2\dfrac {log3}{logx} = 3

    :confused:
    That answers comes out beautifully however the working is long...if you get to the end the answers are x=3 and x=9 , very nice question bro!
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    (Original post by Scottishsiv)
    Anyone who doesn't have one should really buy one, they are only about 10 pounds each on amazon. Although be careful, certain calculators aren't allowed and you should really check the models on edexcel.
    look at page 14 of this document for full details
    http://www.jcq.org.uk/attachments/pu...CE%2009-10.pdf
    hi, just pass by and dropping a msg here. Does my casio fx-9860gii sd allowed in edexcel exam board?
 
 
 
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