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Thread: Functions

  1. #1
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    Functions

    If

    $\displaystyle
    x =\dfrac{e^{2x}}{y^2}
    $

    express z as a function on x and y
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  2. #2
    tah
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    what is the relation between x,y and z?
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  3. #3
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    Im afraid thats all that is in the question! Nothing else - this is a from a sample maths exam that is for interviewing purposes.
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  4. #4
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    there is something missing in your question
    I think it is y^z
    not y^2

    can you check it, please?
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  5. #5
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    Sorry it was indeed a typo in my original post!

    $\displaystyle

    x =\dfrac{e^{2z}}{y^2}
    $
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  6. #6
    tah
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    Quote Originally Posted by tangehayes View Post
    Sorry it was indeed a typo in my original post!

    $\displaystyle

    x =\dfrac{e^{2z}}{y^2}
    $
    Now, it should be easy putting x,y on the same side and using the logarithmic function right!
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  7. #7
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    $\displaystyle
    x =\dfrac{e^{2z}}{y^2}
    $
    then
    $\displaystyle
    x y^2=e^{2z}
    $

    and we know if a=b, then, log a = log b

    $\displaystyle
    log (x y^2)=log (e^{2z})
    $

    can you complete it now by using the logarithmic basic`s laws ?
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  8. #8
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    hmm, I'm a little rusty on my log laws as well. Basically I have an interview next week where I have to do a maths test. I have't used maths in 10 years, but in my time I was quite good at maths, Im afraid Ive forgotten a lot of the principles now. I would really appreciate it.
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  9. #9
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    Quote Originally Posted by tangehayes View Post
    hmm, I'm a little rusty on my log laws as well. Basically I have an interview next week where I have to do a maths test. I have't used maths in 10 years, but in my time I was quite good at maths, Im afraid Ive forgotten a lot of the principles now. I would really appreciate it.

    I'd have taken the previous answerer in base e so that you get
    $\displaystyle \ln({xy^2}) = \ln{e^{2z}}$

    Recalling that $\displaystyle \ln({e^a}) = a$ we get
    $\displaystyle \ln{xy^2} = {2z}$ and then divide through by 2 to give:

    $\displaystyle {z} = \frac{1}{2}\ln({xy^2})$
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