How to prove that a function is a positive montone transformation

Question

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.

Answer 1

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. $\blacksquare$

Answer 2

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