Solving recurrence relation $a_{n+1} = \frac{3a_n^2}{a_{n-1}}$

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I am currently studying recurrence relations and I am stuck at a particular problem:

$$a_{n+1} = \frac{3a_n^2}{a_{n-1}}$$ for $n \geq 1$, starting with $a_0 = 1$, $a_1 = 2$.

In the lecture, we only covered linear recurrence relations, so I am kind of lacking of an ansatz. My idea was making a clever "coordinate" transformation so that the above relation will be simplified. I thought about using something like $\exp$ or $\log$ in order to be able to use addition theorems so that the expression becomes a normal recurrence relation (the square becomes a "$2\times$" and the division a $-$). I am worried about that, because I am only dealing with discrete values.

Therefore, can someone give me a good recommendation for an overview how to deal with this and how I can find a suiting ansatz?

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As you suggested, applying $\log$ on both sides we obtain

$$\log a_{n+1}=2\log a_n-\log a_{n-1}+\log 3.$$