April 2013 Exam Question 5

Discussion in 'SP9' started by kylie jane, Apr 12, 2014.

  1. kylie jane

    kylie jane Member

    Hi,

    Can someone please let me know where I have gone wrong in my calculations below.

    If C(u1,u2,u3,u4) = exp[-{((-lnu1)^a+(-lnu2)^a+(-lnu3)^a+(-lnu4)^a)^(1/a)}]

    C= exp[-{((-1)^a((lnu1)^a+(lnu2)^a+ln(u3)^a+ln(u4)^a))^(1/a)}]
    =exp[-1*(-1)^(a/a)(lnu1*u2*u3*u4)^(a/a)]
    =exp(lnu1*u2*u3*u4)
    =u1*u2*u3*u4

    If what I've done above is correct then I should get the same as the solutions (95.8%) however I get 92.2%. Have I made a mistake somewhere above??
     
  2. r_v.s

    r_v.s Member

    Are you sure that
    [(lnu1)^a + (lnu2)^a+(lnu3)^a+(lnu4)^a] = [ln(u1*u2*u3*u4)]^a
    ????
    I mean in gen, (x^a +y^a) is not (x+y)^ a, isnt it!
     
  3. kylie jane

    kylie jane Member

    Pretty sure as ln(a) + ln(b) = ln(ab) but I actually think my mistake is due to me applying the power laws for logs incorrectly. ln(a^2)=2*ln(a) but I assumed in my workings above that (ln(a))^2= 2ln(a) which is incorrect... So I think that solves it! :)
     
  4. r_v.s

    r_v.s Member

    Thats what i meant! :) ln(a) + ln(b) = ln(ab).
    (ln(a))^c + (ln(b))^c would not add up the same way isnt it!
     
  5. kylie jane

    kylie jane Member

    Yes, but ln(a^c) + ln(b^c) = ln ((ab)^c). Which is what I had assumed in my workings as I included the power within the log. I understand that (a+b)^c does not equal a^c + b^c.
     

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