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    This from the MEI C2 paper from june 2012.

    I basically wrote the 4 logs in exponential form and I got x = 10, 10^5 and y = 10^5, 10^17.
    Then I found the equation, y = 10^12x -(10x10^11).

    Is this miserably wrong?
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    (Original post by UKBrah)
    This from the MEI C2 paper from june 2012.

    I basically wrote the 4 logs in exponential form and I got x = 10, 10^5 and y = 10^5, 10^17.
    Then I found the equation, y = 10^12x -(10x10^11).

    Is this miserably wrong?
    Not quite sure what you've done but it's not right...

    Don't worry about the logs on the axes for the moment: first treat this as any normal straight line and from the two points, work out the gradient, m. Then using y=mx+c, work out the intercept by substituting in one of the points because you have a straight line here. Work out the gradient and intercept and tell me what you get, then we can move on to the next bit.
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    (Original post by Star-girl)
    Not quite sure what you've done but it's not right...

    Don't worry about the logs on the axes for the moment: first treat this as any normal straight line and from the two points, work out the gradient, m. Then using y=mx+c, work out the intercept by substituting in one of the points because you have a straight line here. Work out the gradient and intercept and tell me what you get, then we can move on to the next bit.
    Haha, I got y = 3x + 2.
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    (Original post by UKBrah)
    Haha, I got y = 3x + 2.
    Excellent - that is right. Now replace the "y" by the quantity on the y-axis, and "x" by the quantity on the x-axis.
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    you could start with

    y = Axb

    so

    logy = logA + b(logx).... so the gradient is b and the vertical intercept is logA

    (the base is 10 throughout)...
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    (Original post by Star-girl)
    Excellent - that is right. Now replace the "y" by the quantity on the y-axis, and "x" by the quantity on the x-axis.
    log y = 3logx + 2?
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    And so y =?
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    (Original post by UKBrah)
    log y = 3logx + 2?
    Yes. Now how do you reckon you can simplify this?

    Hint:
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    1. You have two logs which are to the same base, so subtract 3logx from both sides to get it on the same side as logy, then you can use log rules to simplify that side.

    2. Alternatively, take 10^ each side and simplify from there - both methods will get you the same answer.

    Use the method you feel most comfortable with.
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    y = 100x?
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    (Original post by UKBrah)
    y = 100x?
    \log_{10}y = 3\log_{10}x + 2 = \log_{10} 100 x^{3} \Leftrightarrow y = ?
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    \log_{10}y - \log_{10}x^3 = 2

    \log_{10}(\frac{y}{x^3}) = 2

    10^2 = \frac{y}{x^3}

    y=100x^3

    ??
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    (Original post by UKBrah)
    \log_{10}y - \log_{10}x^3 = 2

    \log_{10}(\frac{y}{x^3}) = 2

    10^2 = \frac{y}{x^3}

    y=100x^3

    ??
    Yeah - that's it.
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    (Original post by Star-girl)
    Yeah - that's it.
    wagwan cheers
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    (Original post by UKBrah)
    wagwan cheers
    You're welcome.
 
 
 
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