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    (Original post by mr.holmes)
    but they asked you to find the value for x or what exactly?
    are you sure from your answer?
    i thought we use the y for substituting in the equation
    We could have substituted y or solved the question merely through logarithm laws which is what I did.

    Just think about it if x=81, log3(81) gives 4 which is y. And therefore the expression which Summaiya got for log3 (x/9), y-2, is equals to 2. If you sub x=81 in, you end up with y-2 equaling 2.
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    (Original post by Helper OF ALL)
    so is my answer right....
    Yes
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    (Original post by Helper OF ALL)
    i got y= 1/5 and x= 81...
    and i think these are correct.. becuz i confirmed it with ma fellows
    could you please tell me what the question wanted?
    i actually forgot and i think i did not read it well.
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    (Original post by Sumaiyya1999)
    Yes
    . relief in the A--........
    tom is chem . gotta study it..
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    (Original post by mr.holmes)
    could you please tell me what the question wanted?
    i actually forgot and i think i did not read it well.
    i really cant think about it ... sorry
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    (Original post by Sumaiyya1999)
    O yes but that way way also u should get a quadratic equation cuz (x/9)^2 is x^2/9^2

    Posted from TSR Mobile
    Well I suppose there is that way of solving the question too.
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    (Original post by Fasih178)
    We could have substituted y or solved the question merely through logarithm laws which is what I did.

    Just think about it if x=81, log3(81) gives 4 which is y. And therefore the expression which Summaiya got for log3 (x/9), y-2, is equals to 2. If you sub x=81 in, you end up with y-2 equaling 2.
    so substituting is right?
    and they asked you to find the value for y?
    i think i did everything wrong..........
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    yea first one was y-3
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    (Original post by Helper OF ALL)
    . relief in the A--........
    tom is chem . gotta study it..
    Best of luck
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    anyways, what was the value in the integration question or what was the method you used?
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    (Original post by lympho)
    yea first one was y-3
    y-3 not possible.......because log 9 to the base 3 is 2...

    first was Y-2 and other was Y/2
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    When does the unofficial mark scheme come out?
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    (Original post by mr.holmes)
    anyways, what was the value in the integration question or what was the method you used?
    i used area under curve from 1 to 4 minus area of triangle from x=4 to x=1.8 adn height 11.

    area of triangle = 1/2 * (4 - 1.8) * 11
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    (Original post by mr.holmes)
    so substituting is right?
    and they asked you to find the value for y?
    i think i did everything wrong..........
    Nah man even if you substituted the value of y and ended up with a quadratic equation, you're gonna gain at least 2 marks. The other two marks are for finding a solution of y if you did substitution. But if you solved it through log laws then the remaining two marks are for solution of x instead of y.
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    (Original post by mr.holmes)
    anyways, what was the value in the integration question or what was the method you used?
    5.something, i forget what the value was in fractions
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    (Original post by Helper OF ALL)
    i used area under curve from 1 to 4 minus area of triangle from x=4 to x=1.8 adn height 11.

    area of triangle = 1/2 * (4 - 1.8) * 11
    That's precisely what I did as well.
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    (Original post by Helper OF ALL)
    i used area under curve from 1 to 4 minus area of triangle from x=4 to x=1.8 adn height 11.

    area of triangle = 1/2 * (4 - 1.8) * 11
    I did that too
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    (Original post by Sumaiyya1999)
    5.something, i forget what the value was in fractions
    and what was the method you used?
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    (Original post by lympho)
    When does the unofficial mark scheme come out?
    What is that? Unofficial mark scheme?
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    (Original post by mr.holmes)
    and what was the method you used?
    Same as what fasih and helper of ALL did
 
 
 
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