Consider the utility function

$$ U(x_1,x_2) = x_1^\alpha x_2^\beta $$ for $0 < \alpha, \beta < 1$. How do I then show that $$ V(x_1,x_2) = F(U(x_1,x_2)) = \frac{\alpha}{\beta} \ln(x_1) + \ln(x_2) $$ is a positive, monotone transformation of $U(x_1,x_2)$. I thought about showing that MRS for both functions is the same. Is this approach alright? I also thought about taking $\ln(x)$ on $U(x_1,x_2)$ and see what happens:

$$\ln(x_1^\alpha x_2^\beta) = a \ln(x_1) + \beta \ln(x_2)$$ but I am not sure here if I am allowed to divide with $\frac{1}{\beta}$ now to get the desired result?

Thanks for your help in advance.

  • 4
    $\begingroup$ Well, is multiplication with $1/\beta$ a positive, monotone transformation? $\endgroup$
    – VARulle
    Commented Jun 1, 2020 at 11:17
  • $\begingroup$ I think so, yes. I don't see how it changes anything? Otherwise I am not sure how to get the desired result. $\endgroup$
    – Mathias
    Commented Jun 1, 2020 at 11:46

2 Answers 2


Other people have already hinted at the correct answer. Here is a complete justification:

Note that $V(x,y) = g(U(x,y))$ where $g$ is the function defined by $$g(u) = \dfrac{1}{\beta} \cdot \log(u).$$ Assuming that only positve levels of consumption are allowed (otherwise $V$ is not well defined), $g$ is continuously differentiable on the image of $U$ and satisfies $$ g'(u) = \dfrac{1}{\beta u} > 0 $$ Therefore, $g$ is a monotone transformation. $\blacksquare$


Take the first derivative; if that is always positive for all positive values of your variable then the transformation is positive monotone (i.e maintains ordering)

Since you have two variables, you need to check how the transformation affects both by checking two first order derivatives.

Edit: if you assume the function U is always positive you only need to check the first derivative of the transformation with respect to U

  • $\begingroup$ Why should this depend on the number of variables? There's one transformation, $F(U)=\frac{\ln U}{\beta}$, so there's one derivative to check. $\endgroup$
    – VARulle
    Commented Jun 1, 2020 at 15:33
  • $\begingroup$ There are two transformations, the first is when you take the natural logarithm, the second is when you divide by $\beta$ you need to check that both of these do not change the original ordering of utilities given $x_1$ and $x_2$. Dividing by $\beta$ is a nonissue as long as $\beta$ is positive, it’s just multiplication by a constant. So this boils down to checking the log transformation, which is by definition a positive monotone transformation but to formally check it you need to show that it’s increasing in the values of the original utility function, made up of $x_1$ and $x_2$ $\endgroup$
    – tvbc
    Commented Jun 1, 2020 at 16:09
  • $\begingroup$ You can always view it as one transformation, it won’t really make a difference, however, unless you assume U is always positive you can run into a situation where beta = 1/ln(x2) and x1 is constant...that would not be a positive monotone transformation $\endgroup$
    – tvbc
    Commented Jun 1, 2020 at 16:23

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