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    Hello There, I hope you are well!

    For my second assignment I have been given some maths problems to solve.

    I am finding difficulty in two of the problems:

    The first is

    If x does not equal 2 or 4, verify that

    (x / x-4) + (3 / x-2) = x^2 + x - 12 / (x-4) (x-2)

    - I have managed to do this OK but the second half of this problem states "And hence, or otherwise, find the values of x for which:

    (x / x-4) + (3 / x-2) = 0

    Which has stumped me.

    The second problem is...

    The magnitude of an earthquake is given on the Richter scale by the formula:

    R = log (a / T) + B

    in the usual notation

    A certain earthquake has a magnitude measuring R = 5 on the Richter scale at a particular position where B= 6.8. What is the Richter scale measurement at the same position of an earthquake which is 10 times stronger than this (in the sense that a/T is 10 times its value when R = 5)?

    That has also stumped me.

    Any help/advice would be appreciated!!

    Thank you

    Louk
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    Any help? Would be appreciated from anyone as i truly am stuck.

    Louk
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    (Original post by Louk)
    Hello There, I hope you are well!

    For my second assignment I have been given some maths problems to solve.

    I am finding difficulty in two of the problems:

    The first is

    If x does not equal 2 or 4, verify that

    (x / x-4) + (3 / x-2) = x^2 + x - 12 / (x-4) (x-2)

    - I have managed to do this OK but the second half of this problem states "And hence, or otherwise, find the values of x for which:

    (x / x-4) + (3 / x-2) = 0

    Which has stumped me.

    The second problem is...

    The magnitude of an earthquake is given on the Richter scale by the formula:

    R = log (a / T) + B

    in the usual notation

    A certain earthquake has a magnitude measuring R = 5 on the Richter scale at a particular position where B= 6.8. What is the Richter scale measurement at the same position of an earthquake which is 10 times stronger than this (in the sense that a/T is 10 times its value when R = 5)?

    That has also stumped me.

    Any help/advice would be appreciated!!

    Thank you

    Louk
    For the first one, you know x isn't 2 or 4, so x^2 + x - 12 / (x-4) (x-2) = (x / x-4) + (3 / x-2). Thus if (x / x-4) + (3 / x-2) = 0 this implies x^2 + x - 12 / (x-4) (x-2) = 0 and since the denominator is always positive, x^2 + x - 12 = 0, which can be factorised.

    For the second one, can you tell us what a, T and B represent?
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    that's all i was given for the second question! thanks for your help on the first loads!!

    Louk
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    (Original post by Louk)
    that's all i was given for the second question! thanks for your help on the first loads!!

    Louk
    Ok, say 5 = log(a/T) + 6.8, so log(a/T) = -1.8 so log (10a/T) = log 10 + log (a/T) = log 10 - 1.8 and the value of log 10 depends on your base (i assume you actually mean ln x?)
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    (Original post by theone)
    Ok, say 5 = log(a/T) + 6.8, so log(a/T) = -1.8 so log (10a/T) = log 10 + log (a/T) = log 10 - 1.8 and the value of log 10 depends on your base (i assume you actually mean ln x?)
    no its log to the base 10, thats the one they use for richter.
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    (Original post by elpaw)
    no its log to the base 10, thats the one they use for richter.
    In that case we have log 10 = 1, so the value ought to be -0.8. Thanks elpaw
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    Thank You LOADS!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!

    Louk
 
 
 
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