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C2 Trapezium rule and logs easy questions Watch

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    Trapezium rule:



    For a) I know the height is 0.5, but using the formula h = (b-a)/n I don't get 0.5, I use b as 4 and a as 1 and n as 7, but obviously n has to be 6, thought i see 7 intervals. Is it because the last interval is 0?

    Logs question:



    For part bii) I got the question correct, but I done a complete different method from the mark scheme, and I'm wondering how the mark scheme did it.

    My method:

    

2^x = 5^{x-1}

xlog2=(x-1)log5

\dfrac{x(log2)}{(log5)} = x - 1

x - \dfrac{x(log2)}{(log5)} = 1

x(1 - \dfrac{log2}{log5} ) = 1

\dfrac{1}{1 - \frac{log2}{log5}} = x

    This is correct, but the mark scheme simply got:

    

x = \dfrac{log5}{log5 - log2}

    If anyone could explain how they got this would be great.
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    (Original post by thorn0123)
    Trapezium rule:



    For a) I know the height is 0.5, but using the formula h = (b-a)/n I don't get 0.5, I use b as 4 and a as 1 and n as 7, but obviously n has to be 6, thought i see 7 intervals. Is it because the last interval is 0?

    Logs question:



    For part bii) I got the question correct, but I done a complete different method from the mark scheme, and I'm wondering how the mark scheme did it.

    My method:

    

2^x = 5^{x-1}

xlog2=(x-1)log5

\dfrac{x(log2)}{(log5)} = x - 1

x - \dfrac{x(log2)}{(log5)} = 1

x(1 - \dfrac{log2}{log5} ) = 1

\dfrac{1}{1 - \frac{log2}{log5}} = x

    This is correct, but the mark scheme simply got:

    

x = \dfrac{log5}{log5 - log2}

    If anyone could explain how they got this would be great.

    \dfrac{1}{1 - \frac{log2}{log5}} 



= \dfrac{1}{\frac{log 5 - log2}{log5}}



= \dfrac{log5}{log5 - log2}

    I'll look into the trapezium rule question if it's not answered yet when I have more time. Most likely it will though, answers come fast in this forum.
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    (Original post by thorn0123)
    Trapezium rule:



    For a) I know the height is 0.5, but using the formula h = (b-a)/n I don't get 0.5, I use b as 4 and a as 1 and n as 7, but obviously n has to be 6, thought i see 7 intervals. Is it because the last interval is 0?

    Logs question:



    For part bii) I got the question correct, but I done a complete different method from the mark scheme, and I'm wondering how the mark scheme did it.

    My method:

    

2^x = 5^{x-1}

xlog2=(x-1)log5

\dfrac{x(log2)}{(log5)} = x - 1

x - \dfrac{x(log2)}{(log5)} = 1

x(1 - \dfrac{log2}{log5} ) = 1

\dfrac{1}{1 - \frac{log2}{log5}} = x

    This is correct, but the mark scheme simply got:

    

x = \dfrac{log5}{log5 - log2}

    If anyone could explain how they got this would be great.
    Multiply everything through by log 5 to get rid of the fraction.
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    1-((log2/log5)) = 1/x

    (log5/log5)=1

    (log5-log2)/log5 =1/x

    flip it round to find x (reciprocal)

    log5/(log5-log2) = x

    It's just neater to write this way than yours.

    For trapezium rule question.
    b=4 a=1 n=6 (because there are 6 strips and 7 ordinates) there are always one more ordinate than strips, try drawing it out with 6 strips and you'll see what I mean.
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    Ok I understand the log question, but what is an ordinate? I've drawn 6 strips and I still don't see what you mean.
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    (Original post by thorn0123)
    Ok I understand the log question, but what is an ordinate? I've drawn 6 strips and I still don't see what you mean.
    An ordinate is like a point, there is always one more than the number of strips when using the trapezium rule.

    Edit If you still don't get it
    Spoiler:
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    Ordinates = 7 therefore 6 strips.

    Using  h = \dfrac {b-a}{n}

     \dfrac {4-1}{6} = 0.5
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    the x values are what I meant, so you have 1,1.5,2 to 4.
    7 values (ordinates there)
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    thanks
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    are these from edexcel papers?
 
 
 
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