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

1. ## 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)

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

Are you trying to find w when LD= LS

3. ## 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...

4. ## 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