Are you trying to find w when L^{D}= L^{S}
Your answer is correct for that
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)
lnL^{D }= lnA + aw
lnL^{s}= 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)
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...
To find the equilibrium employment put the expression for w into either of your first equations.
The answer is Employment= ^{b/(b-a)}
or Employment= ^{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 L^{S} and L^{D} against w you first need to get L^{S} on its own from the equation lnL^{S}= lnA +wa
From rules of logs wa= ln(e^{wa})
Then lnL^{S}= lnA +ln(e^{wa})
Combine logs on the right hand side
lnL^{S}= ln(A(e^{wa}))
and you can drop the logs
L^{S}= A(e^{wa})
You can do likewise for L^{D}