Why is the lower bound $m \log n$ for this make-set, union and find-set sequence?

Question

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Is the lower bound $m\log n$ because we are only looking at the lower bound for union by rank only? If we make $n$ MAKE-SET operations, then there would be $\log n$ UNION operations, and then $m - 2n + 1$ FIND-SET operations. The lower bound seems larger to me but what am I missing?