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    (Original post by melinalouise)
    The thing I struggle with most in core 2 is the logarithms questions, anyone have any advice on how to tackle them as they're usually worth lots of marks? oh and I always get the transformations confused too
    I'm completely with you on that, I'm not a fan of them either
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    (Original post by GoldenLotus)
    Jan 2013
    6. a) A geometric series begins 420 + 294 + 205.8
    iii) Write the nth term of the series in the form p x q^n

    As far as I got was Un=420x0.7^(n-1) which got me 1 mark (out of 2)
    so it wants you to write q^n, you have written q^(n-1). The first value is 420, therefore the equation will become. 420=p x 0.7^1. Re-arrange and p=600, so the answer should be un=600 x 0.7^n
    Is that correct?
    Sometimes you need to think less about the equations they give you and logically about the question.
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    (Original post by GoldenLotus)
    Jan 2013
    6. a) A geometric series begins 420 + 294 + 205.8
    iii) Write the nth term of the series in the form p x q^n

    As far as I got was Un=420x0.7^(n-1) which got me 1 mark (out of 2)
    Ok so you were right BUT in this case it was testing your ability to manipulate indices (the question said put it in the form p x q^n NOT p x q^(n-1).

    so have a look at the image below that i have made to explain what to do, excuse that it has been done in paint and let me know if you need further help
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    (Original post by melinalouise)
    The thing I struggle with most in core 2 is the logarithms questions, anyone have any advice on how to tackle them as they're usually worth lots of marks? oh and I always get the transformations confused too
    Logs are difficult but the key is just knowing that you have a toolbelt with about 10 log laws so just always see what the examiner has done to try and muddle up the logs and so just do a small step at a time (e.g. simplify log a - log b to log a/b etc)

    Its difficult to explain without context, so if you were to choose a question/example that you found difficult I would be more than happy to help
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    (Original post by CheaterHater)
    so it wants you to write q^n, you have written q^(n-1). The first value is 420, therefore the equation will become. 420=p x 0.7^1. Re-arrange and p=600, so the answer should be un=600 x 0.7^n
    Is that correct?
    Sometimes you need to think less about the equations they give you and logically about the question.
    Couldn't agree more
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    (Original post by Rager6amer)
    Logs are difficult but the key is just knowing that you have a toolbelt with about 10 log laws so just always see what the examiner has done to try and muddle up the logs and so just do a small step at a time (e.g. simplify log a - log b to log a/b etc)

    Its difficult to explain without context, so if you were to choose a question/example that you found difficult I would be more than happy to help
    This question was from 2014 and I don't know how to do c and d?
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    (Original post by SanaGul)
    Considering the fact that I've studied A2 maths and further maths. I feel quite confident in the C2 paper. The only that still catches me are the Arithmetic and Geometric progressions. Does any body know an effective way of practic8ng these so they fit in my head.
    Try this geometric progression question. It's a question I remember from one of Teeem's threads a while back now.
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    (Original post by melinalouise)
    This question was from 2014 and I don't know how to do c and d?
    Part C: whenever you have a translation, you write the opposite of what you do in into the equation, for example if i have the line y = 6x + 4 and I translate it by vector (-2,0) I have moved it 2 units to the left BUT in the equation I replace 'x' with 'x+2' so therefore the new one would be y = 6(x+2) + 4 so it would become y = 6x +12 +4 so therefore y = 6x +16 (i hope this makes sence)

    So to answer the question in part C you have translated it by (1,p) so remember we write the opposite so therefore y = 3 x 12^x becomes 'y-p = 3 x 12^(x-1)' and so rearrange to get y = 3 x 12^(x-1) + p now since the curve intersects the origin (0,0) we can sub in the values 0 and 0 for x and y so then we get 0 = 3 x 12^(0-1) + p and so therefore 3 x 1/12 + p = 0 so rearrange to get 1/4 + p = 0 and hence p = -1/4. I hope this makes sense.

    As for the logs it would take forever for me to explain so check out the written solution here and see if this makes sense. let me know if you have any more problems http://bit.ly/1RhsgoK
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    (Original post by Rager6amer)
    Part C: whenever you have a translation, you write the opposite of what you do in into the equation, for example if i have the line y = 6x + 4 and I translate it by vector (-2,0) I have moved it 2 units to the left BUT in the equation I replace 'x' with 'x+2' so therefore the new one would be y = 6(x+2) + 4 so it would become y = 6x +12 +4 so therefore y = 6x +16 (i hope this makes sence)

    So to answer the question in part C you have translated it by (1,p) so remember we write the opposite so therefore y = 3 x 12^x becomes 'y-p = 3 x 12^(x-1)' and so rearrange to get y = 3 x 12^(x-1) + p now since the curve intersects the origin (0,0) we can sub in the values 0 and 0 for x and y so then we get 0 = 3 x 12^(0-1) + p and so therefore 3 x 1/12 + p = 0 so rearrange to get 1/4 + p = 0 and hence p = -1/4. I hope this makes sense.

    As for the logs it would take forever for me to explain so check out the written solution here and see if this makes sense. let me know if you have any more problems http://bit.ly/1RhsgoK
    thankyou
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    (Original post by B_9710)
    Try this geometric progression question. It's a question I remember from one of Teeem's threads a while back now.
    Name:  image.jpg
Views: 118
Size:  38.8 KB
    is k=4?
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    (Original post by melinalouise)
    thankyou
    is vector multiplication there in C2
    i know dat translation of f(x) graphs are dere
    bt i have never come accross a vector translation question
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    I would recommend purchasing the Casio fx-991es plus calculator, as it allows you to check the answers to your definite intergrals, definitely worth it!
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    Okay, a logs question... here is the mark scheme for the question;

    Name:  Untitledma.png
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    How do they get from the 2nd line to the third line?

    Thanks in advance
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    (Original post by Blake Jones)
    Okay, a logs question... here is the mark scheme for the question;



    How do they get from the 2nd line to the third line?

    Thanks in advance
    They've literally just factorised 2^x out because it's seen in both of the values (it's a factor of both), if you times the 3rd line out, you will get the 2nd line.
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    (Original post by CheaterHater)
    They've literally just factorised 2^x out because it's seen in both of the values (it's a factor of both), if you times the 3rd line out, you will get the 2nd line.
    Ah I see, thank you, I factorised it wrong, instead of seeing it as squared I accidentally took the 2^x out and then put 2 in the brackets - plonker! haha!! oops! thanks!
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    (Original post by Blake Jones)
    Okay, a logs question... here is the mark scheme for the question;

    Name:  Untitledma.png
Views: 116
Size:  17.5 KB

    How do they get from the 2nd line to the third line?

    Thanks in advance
    Here does this help?
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    (Original post by Rager6amer)
    Here does this help?
    Thanks
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    (Original post by Rager6amer)
    Ok so you were right BUT in this case it was testing your ability to manipulate indices (the question said put it in the form p x q^n NOT p x q^(n-1).

    so have a look at the image below that i have made to explain what to do, excuse that it has been done in paint and let me know if you need further help
    Thank you that helped me so much
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    Think I get it now, thanks to everyone who helped. Can someone please tell me why you can't log individual numbers? Because that's one of my common mistakes, like say if it says 2x^ x 5x = 4x + 6 (random lol I didn't get this from anywhere) why would it be Log(2x^ X 5x) etc? All this time I've been log-ing stuff individually, and I want to fix it 😂
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    Hey. I have done a bunch of questions, and while I did get them all correct, I used a different method to what is specified in the mark scheme (I used the binomial formula method and made it (1+ax)^n rather than nCr like is taught at C2). Would I still get full marks for this or would I lose the working marks?
 
 
 
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