0
$\begingroup$

If the utility is $U = \ln x + 2 \ln y$, how do you compute Walrasian equilibrium via usual formula for demand $x=a(x p_x + y p_y)/p_x(a+b)$ ?

What is $a$ and $b$?

In case of Cobb-Douglas function like $U=x^3y^4$ it would be simple: $a=3$ and $b=4$, but in this case how is it computed?

$\endgroup$
1
$\begingroup$

let $u \equiv xy^2$, then we have: $$U=lnx+2lny=ln(xy^2)=ln(u)$$ Since $U'(u)>0 \: \forall u>0$ it follows that the $(x,y)$ that maximizes $U$ also maximizes $u$; $\max \{U\}=\max\{ln(u)\}=ln(\max\{u\})$. $u$ represents the same preferences as $U$.

Clearly, $u(x,y)=x^a y^b$ with $a=1$, $b=2$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ So it's the antilogarithm, thank you. That's what happens when you spend too much time on economics stackexchange instead of math. $\endgroup$ – Svit Valenčič Jan 25 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.