How to prove that a function is a positive montone transformation
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. $\blacksquare$