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    (Original post by LastMinReviseGuy)
    Who told you i had a big willy?
    Sorry , but i don't.
    pls... stahp
    (Original post by natninja)
    There are some core 4 papers where the A boundary is significantly higher. The one I say had an A boundary at 70 and the one the following January (yes this was way back in the past) had an A boundary of 71 and an A* boundary of 73...
    these are ones from around 2005 to todays date
    https://c4a3f001dcd45afe69d0ceec8300...%20A-level.pdf

    i don't see any with those sort of boundaries for an A
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    (Original post by thefatone)
    pls... stahp


    these are ones from around 2005 to todays date
    https://c4a3f001dcd45afe69d0ceec8300...%20A-level.pdf

    i don't see any with those sort of boundaries for an A
    In your link, C4 Jan 11 it says it's 69 for an A.
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    (Original post by Zacken)
    In your link, C4 Jan 11 it says it's 69 for an A.
    O.O i seemed to have glanced over that one.... oh well...

    can you help me with this actual logs question then?

    given that
    t=log_3 x
    find in terms of t

    log_3 x^2





log_9 x
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    (Original post by thefatone)
    O.O i seemed to have glanced over that one.... oh well...

    can you help me with this actual logs question then?

    given that
    t=log_3 x
    find in terms of t

    log_3 x^2



log_9 x
    Do you know that \log_nx^m=m\log_nx?

    And that \log_nx=\dfrac{\log_mx}{\log_mn}?

    And that \log_nn=1?
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    (Original post by Kvothe the arcane)
    Do you know that \log_nx^m=m\log_nx?

    And that \log_nx=\dfrac{\log_mx}{\log_mn}?

    And that \log_nn=1?
    Ah thanks, i got the right answer but i'm not sure how i reached it xD
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    (Original post by thefatone)
    Ah thanks, i got the right answer but i'm not sure how i reached it xD
    Type out your working?
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    (Original post by Kvothe the arcane)
    Type out your working?
    ok so  log_9 x = \dfrac{log_3 x}{log_3 9} = \dfrac{log_3 x}{2}= \dfrac{t}{2}


    next question is hence or otherwise find to 3 sig figs the value of x such that

    log_3 x^2 - log_9 x = 4
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    (Original post by thefatone)
    ok so  log_9 x = \dfrac{log_3 x}{log_3 9} = \dfrac{log_3 x}{2}= \dfrac{t}{2}


    next question is hence or otherwise find to 3 sig figs the value of x such that

    log_3 x^2 - log_9 x = 4
    Your working above seems fine. Solve for t and then solve for x .
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    (Original post by Kvothe the arcane)
    Your working above seems fine. Solve for t and then solve for x .
    what if it said express

     log_3 2x

    in terms of t how would i do this?
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    (Original post by thefatone)
    what if it said express

     log_3 2x

    in terms of t how would i do this?
    \log_3 2 + \log_3 x
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    (Original post by Zacken)
    \log_3 2 + \log_3 x
    thank you

    but what if it says something like

    log_3 2x+1

    what do i do then?
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    (Original post by thefatone)
    thank you

    but what if it says something like

    log_3 2x+1

    what do i do then?
    What's the difference? 1 + \log_3 2 + \log_3 x
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    (Original post by Zacken)
    What's the difference? 1 + \log_3 2 + \log_3 x
    do the brackets make a difference?

    log_3 \left(2x+1\right)?
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    (Original post by thefatone)
    do the brackets make a difference?

    log_3 \left(2x+1\right)?
    Yes, obviously so. In that case, it's not representable in terms of t.
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    (Original post by Zacken)
    Yes, obviously so. In that case, it's not representable in terms of t.
    is it not?

    can't you just split it up? or does that not work?
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    (Original post by thefatone)
    is it not?

    can't you just split it up? or does that not work?
    How do you propose on splitting it up...?
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    (Original post by Zacken)
    How do you propose on splitting it up...?
    hmmm ah i see, the +1 inside with the 2x means you can't do anything about it
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    (Original post by thefatone)
    hmmm ah i see, the +1 inside with the 2x means you can't do anything about it
    Yeah.
 
 
 
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