Combine, Then Undo

$\log_{6}\left(x\right)+\log_{6}\left(x-1\right) = 1$ has two logs, and the undo move works on one. So join them first with the product rule, which holds while both inputs are positive: $\log_{6}\left(x\right)+\log_{6}\left(x-1\right) = \log_{6}\left(x\left(x-1\right)\right)$.

Now there is one log, and raising 6 to both sides undoes it: $x\left(x-1\right) = 6$, an ordinary quadratic. Expand, move the 6 across and factor: $\left(x-3\right)\left(x+2\right) = 0$, so x is 3 or -2.

Those are only candidates. Put each back into the original equation. With x = 3 the inputs are 3 and 2, both positive, and $\log_{6}\left(3\right)+\log_{6}\left(2\right)$ is $\log_{6}\left(6\right)$, which is 1. With x = -2 the first input is -2, and the log of a negative number has no value, so -2 is thrown out. The board's move Check x = ... in the original equation runs this test for each candidate.