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    can someone please me where I went wrong on this question, in the solutionbank they sub in the conditions half way through the working out, i'm not sure why my method isn't valid?

    the answer is meant to be (8+4y)/(2-y)
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    (Original post by Katiee224)
    can someone please me where I went wrong on this question, in the solutionbank they sub in the conditions half way through the working out, i'm not sure why my method isn't valid?

    the answer is meant to be (8+4y)/(2-y)
    when you removed the logs on the first page near the end you added the terms when really the constant is multiplied by the other term
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    (Original post by Katiee224)
    can someone please me where I went wrong on this question, in the solutionbank they sub in the conditions half way through the working out, i'm not sure why my method isn't valid?

    the answer is meant to be (8+4y)/(2-y)
    In the line where you went from = \ln |\sec x| + c to  = \ln |\sec x| + \ln A you then left it as a + A when instead you should have done \ln |\sec x| + \ln A = \ln |A\sec x|

    This makes your solution \displaystyle \sqrt{\frac{2+y}{2-y}} = A\sec x \neq \sec x + A. You capiche?
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    ...damn, got ninja-ed there. Oh well, hope you don't mind my un-needed contribution. :-)
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    (Original post by Zacken)
    ...damn, got ninja-ed there. Oh well, hope you don't mind my un-needed contribution. :-)
    Yours was a more detailed explanation.
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    (Original post by Zacken)
    damn, got ninja-ed there.
    One of the reasons I reduced my level of replies - PRSOM.
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    (Original post by 13 1 20 8 42)
    when you removed the logs on the first page near the end you added the terms when really the constant is multiplied by the other term
    (Original post by Zacken)
    In the line where you went from = \ln |\sec x| + c to  = \ln |\sec x| + \ln A you then left it as a + A when instead you should have done \ln |\sec x| + \ln A = \ln |A\sec x|

    This makes your solution \displaystyle \sqrt{\frac{2+y}{2-y}} = A\sec x \neq \sec x + A. You capiche?
    thank youuuuuuuuu i see where I went wrong now

    but out of interest, why have I got to combine the In(secx) with InA before I can remove the logs? have all the log terms got to be grouped up before you can then move the log part?
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    (Original post by Katiee224)
    thank youuuuuuuuu i see where I went wrong now

    but out of interest, why have I got to multiply the In(secx) with InA before I can remove the logs? have all the log terms got to be grouped up before you can then move the log part?
    Well basically what you're doing when you remove the logs is applying e to the power of each side. You technically don't have to explicitly write the log terms grouped up first, but you have to know what will happen to them when you make this manipulation.
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    (Original post by Katiee224)
    thank youuuuuuuuu i see where I went wrong now

    but out of interest, why have I got to multiply the In(secx) with InA before I can remove the logs? have all the log terms got to be grouped up before you can then move the log part?
    Yes, in general when you have something like \ln A = \ln B +c then you're taught to group everything up as \ln A + \ln Be^c \Rightarrow A = Be^c. The real reason why you're doing this is because you're taking the \exp (exponential) of both sides of the equation.

    So it's a bit like when you have x^3 = a + b \Rightarrow x = (a+b)^{1/3} \neq a^{1/3} + b^{1/3}, here you take the exponential of both sides to get:

    \ln A = \ln B + c \Rightarrow e^{\ln A} = e^{\ln B + c} = e^{\ln B}e^c \Rightarrow A = Be^c using the fact that e^{\ln x} = x.
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    (Original post by 13 1 20 8 42)
    Well basically what you're doing when you remove the logs is applying e to the power of each side. You technically don't have to explicitly write the log terms grouped up first, but you have to know what will happen to them when you make this manipulation.
    I give up. :rofl:
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    (Original post by Zacken)
    I give up. :rofl:
    don't feel too bad...it's the last day of term at uni, it's my birthday, and here I am replying as fast as possible to maths queries..
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    (Original post by 13 1 20 8 42)
    don't feel too bad...it's the last day of term at uni, it's my birthday, and here I am replying as fast as possible to maths queries..
    HAPPY BIRTHDAY!!! :party::party:
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    (Original post by 13 1 20 8 42)
    Well basically what you're doing when you remove the logs is applying e to the power of each side. You technically don't have to explicitly write the log terms grouped up first, but you have to know what will happen to them when you make this manipulation.
    (Original post by Zacken)
    Yes, in general when you have something like \ln A = \ln B +c then you're taught to group everything up as \ln A + \ln Be^c \Rightarrow A = Be^c. The real reason why you're doing this is because you're taking the \exp (exponential) of both sides of the equation.

    So it's a bit like when you have x^3 = a + b \Rightarrow x = (a+b)^{1/3} \neq a^{1/3} + b^{1/3}, here you take the exponential of both sides to get:

    \ln A = \ln B + c \Rightarrow e^{\ln A} = e^{\ln B + c} = e^{\ln B}e^c \Rightarrow A = Be^c using the fact that e^{\ln x} = x.
    thank you so much both of you!! makes sense to me now:cute:

    zacken in there with a very detailed answer, you should be teaching maths you are really good at explaining things
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    (Original post by Zacken)
    HAPPY BIRTHDAY!!! :party::party:
    happy birthdayyyyy
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    (Original post by Zacken)
    HAPPY BIRTHDAY!!! :party::party:
    haha thanks. I apologise for my peremptory ninja's...
 
 
 
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