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    Hi!

    I was doing a C2 past paper and came across this question - it seems to involve working backwards from a log graph. The only problem is I don't really understand how to do it...

    I got as far as finding the gradient as 3. I'm not entirely sure what to do after that so if someone could help me through it i'd appreciate it.
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    (Original post by Avacado)
    Hi!

    I was doing a C2 past paper and came across this question - it seems to involve working backwards from a log graph. The only problem is I don't really understand how to do it...

    I got as far as finding the gradient as 3. I'm not entirely sure what to do after that so if someone could help me through it i'd appreciate it.
    This is in the form \log_{10} y = m\log_{10}x + c \Rightarrow Y =mX +c (i.e: it's a linear-log graph)

    You know the gradient is 3, so m=3 - can you find the value of c by plugging a point in? I'll get you started:

    5 = 3(1) + c \Rightarrow c = ?

    Then, you know \log_{10} y = 3 \log_{10}x + c = \log_{10}x^3 + c \Rightarrow y = 10^c \times x^3.
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    so y = 3x + 2

    log(y) = log(x^3) + 2

    y = 10^3 x^3

    y = 1000x^3

    Thanks.. but two questions if I may:

    1) Where does the +2 go?
    2) How do I know if a graph is of the form log(y) = klog(x) + c
    or just
    log(y) = kx + c

    Assuming I'm not given the axis in the first place? I've always been slightly confused about that.
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    (Original post by Avacado)
    so y = 3x + 2

    log(y) = log(x^3) + 2

    y = 10^3 x^3

    y = 1000x^3

    Thanks.. but two questions if I may:
    That's not correct. We have \log_{10} y = \log_{10} x^3 + 2 = \log_{10} x^3 + \log_{10} 10^2 = \log_{10} 100x^3

    So "cancelling" the logs, we have: y = 100x^3.

    1) Where does the +2 go?
    As above. It shouldn't have gone.

    2) How do I know if a graph is of the form log(y) = klog(x) + c
    or just
    log(y) = kx + c
    You will be told by either axis labels or "this is a graph of \log y v/s x" in which case your graph is \log y = k x + c or "this is a graph of \log y v/s \log x" in which case your graph is \log y = k \log x + c . You getting it?


    Assuming I'm not given the axis in the first place? I've always been slightly confused about that.
    By the way, quote me by clicking the reply button at the bottom right of my post so I get a notification.
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    (Original post by Zacken)
    That's not correct. We have \log_{10} y = \log_{10} x^3 + 2 = \log_{10} x^3 + \log_{10} 10^2 = \log_{10} 100x^3

    So "cancelling" the logs, we have: y = 100x^3.



    As above. It shouldn't have gone.



    You will be told by either axis labels or "this is a graph of \log y v/s x" in which case your graph is \log y = k x + c or "this is a graph of \log y v/s \log x" in which case your graph is \log y = k \log x + c . You getting it?




    By the way, quote me by clicking the reply button at the bottom right of my post so I get a notification.
    Ah right that makes a lot more sense. Thank you! I guess I just need to learn the relationship between y=mx+c and the log graphs... you'd think doing physics would make that simple.
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    (Original post by Avacado)
    Ah right that makes a lot more sense. Thank you! I guess I just need to learn the relationship between y=mx+c and the log graphs... you'd think doing physics would make that simple.
    All you have to do when you a straight line graph is write down Y = mX + c, with the capitals and then say \log y = Y and X = x or X = \log x or whatever you're given and all will be clear from there.

    Glad it helped.
 
 
 
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