Big O of multiple variables in a logarithm: is $(2n+m)\log(2n+m) = O((n+m)\log(n+m))$?
Classical way is same as when we consider double limit $\lim\limits_{(n,m) \to \infty}f(n,m)=A$, which use intention to infinity by squares  i.e. for $\forall \varepsilon&gt;0$ $\exists \delta&gt;0$ such that for $\forall n &gt; \delta$ and $\forall m &gt; \delta$ holds $|f(n,m)-A|&lt; \varepsilon$. Using maximum norm conception we can write $\left\Vert(m,n)\right\Vert = \max\{n,m \}&gt; \delta$. It is also known Chebyshev norm and the infinity norm terms for this conception and notation $\left\Vert \cdot \right\Vert_\infty$. Let me say, that this is same as intention to infinity by rectangles i.e. when we require $\exists \delta_1&gt;0, \exists \delta_2&gt;0$ and consider $\forall n &gt; \delta_1$ and $\forall m &gt; \delta_2$.
Applying to big-$O$ we can write that $O(f(n,m)),(n,m) \to \infty$ is set of functions $g$, such, that $\exists C_g&gt;0$ and $\exists M_g&gt;0$ such that for $\forall \left\Vert  (n,m)\right\Vert_\infty &gt; M_g$ holds $g(n,m) \leqslant C_g f(n,m)$.
In your case $(2n+m)\log(2n+m) \leqslant 2(n+m)\log [2(n+m)] \leqslant 4(n+m)\log (n+m)$ using $1+\frac{\log 2}{\log (n+m)} \leqslant 2$, when $n&gt;1,m&gt;1$.
For completeness let me also note, that definition from one variable can be expanded to two or multi  variables in different ways, for example, we can consider intention to infinity by circles i.e. when we consider $\sqrt{n^2+m^2} \to \infty$. Is known intention to infinity by triangles i.e. $n+m \to \infty $ etc.