Given the equation: (3 x)2 - 5 3 x + 6 = 0.

Here's the solution to the problem:

Given the equation: $(\log_3 x)^2 - 5 \log_3 x + 6 = 0$.

Step 1: Let $y = \log_3 x$. Substitute $y$ into the given equation to form a quadratic equation. $y^2 - 5y + 6 = 0$

Step 2: Factor the quadratic equation. We need two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. $(y - 2)(y - 3) = 0$

Step 3: Solve for the possible values of $y$. $y - 2 = 0 \quad or \quad y - 3 = 0$ $y = 2 \quad or \quad y = 3$

Step 4: Substitute back $\log_3 x$ for $y$ and solve for $x$ in each case. Case 1: $y = 2$ $\log_3 x = 2$ Convert the logarithmic equation to an exponential equation: $x = 3^2$. $x = 9$

Case 2: $y = 3$ $\log_3 x = 3$ Convert the logarithmic equation to an exponential equation: $x = 3^3$. $x = 27$

The possible values of $x$ are $\boxed{9}$ and $\boxed{27}$.

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