How to prove that a function is a positive montone transformation Consider the utility function

$$
U(x_1,x_2) = x_1^\alpha x_2^\beta
$$
for $0 &lt; \alpha, \beta &lt; 1$. 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.
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
$$
g'(u) = \dfrac{1}{\beta u} &gt; 0
$$
Therefore, $g$ is a monotone transformation. $\blacksquare$