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. (Sweating)

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)

Re: Economics Algebra - Not sure if this is the right place for this?

Are you trying to find w when L^{D}= L^{S}

Your answer is correct for that

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...

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= $\displaystyle B(A/B)$^{b/(b-a)}

or Employment= $\displaystyle A(A/B)$^{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 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}