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Math Help - Economics Algebra - Not sure if this is the right place for this?

  1. #1
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    Economics Algebra - Not sure if this is the right place for this?

    Hello,

    I'm new to this forum and if I end up posting something that's against the rules, please tell me! I wasn't really sure where to post this but it's just algebra... I think.

    Anyways, I'm currently studying labour economics and my math isn't too great... I was wondering if anybody could help me with this supposedly simple algebraic question.

    So basically I have two equations: (labour supply and demand)

    lnLD = lnA + aw
    lnLs= lnB + bw

    I need to algebraically solve for the equilibrium wage!

    I tried doing it and I ended up something that looked like this (which I think is really off):
    w = ln(A/B) / (b-a)
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  2. #2
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    Re: Economics Algebra - Not sure if this is the right place for this?

    Are you trying to find w when LD= LS
    Your answer is correct for that
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  3. #3
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    Re: Economics Algebra - Not sure if this is the right place for this?

    Hello,

    Yes I am!

    Thank you. If that is correct, do you know how I would find the equilibrium employment level since now I have the equilibrium wage?

    I'm also trying to graph this as well...
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    Re: Economics Algebra - Not sure if this is the right place for this?

    To find the equilibrium employment put the expression for w into either of your first equations.
    The answer is Employment= B(A/B)b/(b-a)
    or Employment= A(A/B)a/(b-a)
    Depending on which equation you put the expression for w into. They are actually equal although they are 2 quite different ways of expressing employment.

    To graph LS and LD against w you first need to get LS on its own from the equation lnLS= lnA +wa
    From rules of logs wa= ln(ewa)
    Then lnLS= lnA +ln(ewa)
    Combine logs on the right hand side
    lnLS= ln(A(ewa))
    and you can drop the logs
    LS= A(ewa)

    You can do likewise for LD
    Thanks from alpinion
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