For each $n \ge 1$, let $P(n)$ be the statement to be proven. We will argue by induction. Indeed,

* Base step: For $n=1$, consider $3^1=3$ coins, and without loss of generality, suppose that two are black and one is white. Put a black on each pan and set the white aside. If the scale balances, then white coin is counterfeit, and since it is white, is lighter than the other coins. If the scales tips, say the left side down, then the black on the left is heavy and fake. In any case, the counterfeit coin is identified from among three coins and classified as heavy or light, so $P(1)$ holds.

* Next we consider the case when $n=3$, assuming, for the moment, that $P(2)$ has been shown. Consider $3^3=27$ coins, with, say, 14 black, and 13 white. Partition these 27 coins into three groups, say $G_1 = (4B, 5W)$, $G_2 = (5B, 4W)$, and $G_3 = (5B, 4W)$. Put $G_2$ on the left, weighed against $G_3$ on the right. If the scale balances, the counterfeit coin is in group $G_1$ and so $P(2)$ applies to a set of 9 coins, 4 black and 5 white. If the left pan goes down, either one of the 5 black from $G_2$ is heavy, or one of the 4 white from $G_3$ is light. The induction hypothesis $P(2)$ now applies to these $3^2 = 9$ coins. Similarly, if the scale tips to the right, the counterfeit is among the 4 white in $G_2$ or the 5 black in $G_3$, and again $P(2)$ applies.

* Inductive step: Fix $k \ge 1$ and assume that $P(k)$ is true. Consider a collection of $3^{k+1}$ coins, one of which is counterfeit. Assume that there is one more black than white (the same argument works if there is one more white than black), so suppose that $\frac{3^{k+1}+1}{2}$ are black and $\frac{3^{k+1}-1}{2}$ are white. Partition the coins into three groups, $G_1, G_2$, and $G_3$, each with $3^k$ coins, and each with a near balance of black and white: In $G_1$, put $\frac{3^k-1}{2}$ black and $\frac{3^k+1}{2}$ white coins. In both $G_2$ and $G_3$, put $\frac{3^k+1}{2}$ black and $\frac{3^k-1}{2}$ white coins. Putting $G_1$ aside, weigh $G_2$ on the left against $G_3$ on the right. If the scales balance, the counterfeit coin is in $G_1$, and $P(k)$ applies to $G_1$, finding the fake coin in an additional $k$ weighings, $k+1$ in all, and so $P(k+1)$ is true in this case. If the scales go down on the left, either one of the $\frac{3^k+1}{2}$ black coins from $G_2$ is heavy, or one of the $\frac{3^k-1}{2}$ white coins from $G_3$ is light and these
$$\frac{3^k + 1}{2} + \frac{3^k - 1}{2} = 3^k$$
coins satisfy the hypothesis for $P(k)$, and so the coin is found in $k+1$ weighings. The analogous argument works when the right pan lowers.

Finally, by mathematical induction, for each $n \ge 1$, we conclude that $P(n)$ holds.