ThatWasHard!
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#1
Report Thread starter 7 years ago
#1
I was wondering how you did question 6 from the june 2012 MEI OCR c2 paper?

I got to the gradient part (3), but didn't really know what to do from there...

I checked the mark scheme, but I don't really udnerstand it...

Thank you

link: http://www.ocr.org.uk/Images/131161-...athematics.pdf
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Smaug123
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#2
Report 7 years ago
#2
(Original post by ThatWasHard!)
I was wondering how you did question 6 from the june 2012 MEI OCR c2 paper?

I got to the gradient part (3), but didn't really know what to do from there...

I checked the mark scheme, but I don't really udnerstand it...

Thank you

link: http://www.ocr.org.uk/Images/131161-...athematics.pdf
OK, so you found the gradient of this line. That is, you found that \log_{10}(y) = 3 \log_{10}(x) + c, for some c. Can you find c now?

EDIT: Not sure where you got 3 from, actually - how did you find that value? (It's not right.)
DOUBLE EDIT: Ignore me, I'm a fool.
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ThatWasHard!
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#3
Report Thread starter 7 years ago
#3
(Original post by Smaug123)
OK, so you found the gradient of this line. That is, you found that \log_{10}(y) = 3 \log_{10}(x) + c, for some c. Can you find c now?

EDIT: Not sure where you got 3 from, actually - how did you find that value? (It's not right.)
DOUBLE EDIT: Ignore me, I'm a fool.
So here's where i'm stuck... Do I use (1,5) to give me:
log(10)(5) = 3log(10)(1)+c?
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brianeverit
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#4
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#4
(Original post by ThatWasHard!)
So here's where i'm stuck... Do I use (1,5) to give me:
log(10)(5) = 3log(10)(1)+c?
Yes. But now you have to make y the subject.
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ThatWasHard!
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#5
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#5
(Original post by brianeverit)
Yes. But now you have to make y the subject.
But sure 3log(10)(1) = 0, as log(10)(1) = 0

How do you make y the subject in this case?
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BabyMaths
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#6
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#6
(Original post by ThatWasHard!)
But sure 3log(10)(1) = 0, as log(10)(1) = 0

How do you make y the subject in this case?
You have found c. Put it back in your equation and get rid of the logs.
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brianeverit
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#7
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#7
(Original post by ThatWasHard!)
But sure 3log(10)(1) = 0, as log(10)(1) = 0

How do you make y the subject in this case?
Combine the two logarithmic terms using the usual laws of logarithms. Then when you have something like \log y=x convert to exponential form, i.e. y=e^x
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