<p>Note that <span class="math-container">$\exp(\ln(x)\ln(y))\ne \exp(\ln(x+y))$</span>. Instead,
<span class="math-container">\begin{equation}
\exp(\ln(x)\ln(y))=x^{\ln y}=y^{\ln x}. 
\end{equation}</span>
The MRS of the monotonically transformed utility is still <span class="math-container">$\frac{y\ln y}{x\ln x}$</span>.</p>

CC BY-SA 4.0
2021-11-15T02:27:27.140Z
2021-11-15T02:27:27.140Z
Herr K.
<p><em>Consider U(x, y) = ln(x)ln(y). Is it quasi-linear?</em></p>
<p>My response was no, since applying an exponential would yield exp(ln(x)ln(y))= exp(ln(x+y)) = x+y.</p>
<p>However, the original MRS was (y/x)(ln(y)/ln(x)). Now it's 1. What did I do wrong?</p>
<p>Thanks in advance!</p>

CC BY-SA 4.0
2021-11-14T19:10:10.697Z
Are you sure that $\exp(\ln(x)\ln(y))= \exp(\ln(x+y))$?
Herr K.
CC BY-SA 4.0
2021-11-14T21:28:52.923Z
@HerrK. That's an answer (:
Giskard
<consumer-theory>
MRS not invariant under monotonic transformation
Renan Andrade
# MRS not invariant under monotonic transformation

(Asked by: username_0 on 2021-11-14)

*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!

**Tags:** consumer-theory

> **Comments:**
> **username_1**: Are you sure that $\exp(\ln(x)\ln(y))= \exp(\ln(x+y))$?  
> **username_2**: @HerrK. That's an answer (:  

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## 1 answers

### Answer by username_1 on 2021-11-15. Score: 3 | Accepted answer

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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