Consider U(x, y) = ln(x)ln(y). Is it quasi-linear?

My response was no, since applying an exponential would yield exp(ln(x)ln(y))= exp(ln(x+y)) = x+y.

However, the original MRS was (y/x)(ln(y)/ln(x)). Now it's 1. What did I do wrong?

Thanks in advance!

  • 1
    $\begingroup$ Are you sure that $\exp(\ln(x)\ln(y))= \exp(\ln(x+y))$? $\endgroup$
    – Herr K.
    Nov 14 '21 at 19:10
  • 1
    $\begingroup$ @HerrK. That's an answer (: $\endgroup$
    – Giskard
    Nov 14 '21 at 21:28

Note that $\exp(\ln(x)\ln(y))\ne \exp(\ln(x+y))$. Instead, \begin{equation} \exp(\ln(x)\ln(y))=x^{\ln y}=y^{\ln x}. \end{equation} The MRS of the monotonically transformed utility is still $\frac{y\ln y}{x\ln x}$.


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