<p>Note that
<span class="math-container">$$ln(x_1)+2*ln(x_2) = ln(x_1)+ln(x_2^2) = ln (x_1 * x_2^2),$$</span>
and note that
<span class="math-container">$$MRS_v = \frac{g'(x_1 * x_2^2) x_1 * 2 x_2}{g'(x_1 * x_2^2)   x_2^2}$$</span>
such that the derivative of <span class="math-container">$g$</span> cancels out.</p>
<p>For more intuition, see <a href="https://economics.stackexchange.com/q/5044/11590">here.</a></p>

CC BY-SA 4.0
2020-11-26T16:53:02.893Z
Nice that makes sense. Thank you
Lifeni
CC BY-SA 4.0
2020-11-26T16:55:28.927Z
You should also see that if you prefer bundle x1,x2 to y1,y2 then it must be that u(x1,x2) > u(y1,y2). If you now apply a positive monotone transformation g, you can see that u(x1,x2) > u(y1,y2) if and only if g(u(x1,x2)) > g(u(y1,y2)).
Bayesian
CC BY-SA 4.0
2020-11-26T16:56:45.313Z
@Lifeni if you think this answer answered your question consider accepting it.
1muflon1
CC BY-SA 4.0
2020-11-26T17:06:09.820Z
Hi Bayesian that makes sense and the intuition in the link makes sense but have can I formally show that the ranking  of the two indifference curves is the same when () is monotonous?
Lifeni
CC BY-SA 4.0
2020-11-26T17:17:09.250Z
See my comment if you compare any two bundles then you have u(x1,x2) > u(y1,y2) if and only if g(u(x1,x2)) > g(u(y1,y2)) by definition of g being a positive monotone transformation. Therefore the ranking is not changed.
Bayesian
CC BY-SA 4.0
2020-11-26T16:47:02.400Z
2020-11-26T16:47:02.400Z
Bayesian
<p>We let <span class="math-container">$g(z)$</span> be a strictly monotonous function so:
<span class="math-container">$$\frac{dg(z)}{dz}&gt;0$$</span>
Consumer 1 has preferences given by the utility function <span class="math-container">$u(x_1,x_2)=ln(x_1)+2ln(x_2)$</span>, while consumer 2 has preferences given by n <span class="math-container">$v(x_1,x_2)=g(x_1x_2^2)$</span>.
Then I have to show that consumer 2 got same preferences as consumer 1.
I think I have to use MSR on <span class="math-container">$x_1x_2^2$</span> and on <span class="math-container">$v(x_1,x_2)$</span>. For MSR on <span class="math-container">$x_1x_2^2$</span>  I get:
<span class="math-container">$$MRS=-\frac{\frac{\partial }{\partial x_1}}{\frac{\partial }{\partial x_2}}=-\frac{x_2^2}{2x_1x_2}$$</span>
But How can I find MSR on <span class="math-container">$v(x_1,x_2)$</span> (I think I got the same MRS if I use the chain rule, but I'm not sure?), and how can I use this to conclude that the ranking of the two indifference curves is the same when <span class="math-container">$g(z)$</span> is monotonous?
I hope that someone can help me?</p>

CC BY-SA 4.0
2020-11-26T16:41:26.087Z
Note that ln(x1)+2*ln(x2) = ln(x1)+ln(x2^2) = ln (x1 * x2^2) and then verify that MRS is invariant to any positive monotone transforation.
Bayesian
<microeconomics><utility><cobb-douglas><ranking>
Utility functions and positive monotone transformations
Lifeni
# Utility functions and positive monotone transformations

(Asked by: username_0 on 2020-11-26)

We let $g(z)$ be a strictly monotonous function so:
$$\frac{dg(z)}{dz}>0$$
Consumer 1 has preferences given by the utility function $u(x\_1,x\_2)=ln(x\_1)+2ln(x\_2)$, while consumer 2 has preferences given by n $v(x\_1,x\_2)=g(x\_1x\_2^2)$.
Then I have to show that consumer 2 got same preferences as consumer 1.
I think I have to use MSR on $x\_1x\_2^2$ and on $v(x\_1,x\_2)$. For MSR on $x\_1x\_2^2$ I get:
$$MRS=-\frac{\frac{\partial }{\partial x\_1}}{\frac{\partial }{\partial x\_2}}=-\frac{x\_2^2}{2x\_1x\_2}$$
But How can I find MSR on $v(x\_1,x\_2)$ (I think I got the same MRS if I use the chain rule, but I'm not sure?), and how can I use this to conclude that the ranking of the two indifference curves is the same when $g(z)$ is monotonous?
I hope that someone can help me?

**Tags:** microeconomics, utility, cobb-douglas, ranking

> **Comments:**
> **username_1**: Note that ln(x1)+2*ln(x2) = ln(x1)+ln(x2^2) = ln (x1 * x2^2) and then verify that MRS is invariant to any positive monotone transforation.  

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## 1 answers

### Answer by username_1 on 2020-11-26. Score: 2 | Accepted answer

Note that
$$ln(x\_1)+2\*ln(x\_2) = ln(x\_1)+ln(x\_2^2) = ln (x\_1 \* x\_2^2),$$
and note that
$$MRS\_v = \frac{g'(x\_1 \* x\_2^2) x\_1 \* 2 x\_2}{g'(x\_1 \* x\_2^2) x\_2^2}$$
such that the derivative of $g$ cancels out.

For more intuition, see [here.](https://economics.stackexchange.com/q/5044/11590)

> **Comments:**
> **username_0**: Nice that makes sense. Thank you  
> **username_1**: You should also see that if you prefer bundle x1,x2 to y1,y2 then it must be that u(x1,x2) > u(y1,y2). If you now apply a positive monotone transformation g, you can see that u(x1,x2) > u(y1,y2) if and only if g(u(x1,x2)) > g(u(y1,y2)).  
> **username_2**: @username_0 if you think this answer answered your question consider accepting it.  
> **username_0**: Hi username_1 that makes sense and the intuition in the link makes sense but have can I formally show that the ranking  of the two indifference curves is the same when () is monotonous?  
> **username_1**: See my comment if you compare any two bundles then you have u(x1,x2) > u(y1,y2) if and only if g(u(x1,x2)) > g(u(y1,y2)) by definition of g being a positive monotone transformation. Therefore the ranking is not changed.  

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