Solve the equation: (3 x)2 - (1)/(2) 3 x = (3)/(2)

Substitute this into the given equation.
The equation is $(\log_3 x)^2 - \frac{1}{2} \log_3 x = \frac{3}{2}$.
Substituting $y = \log_3 x$, we get:

$$y^2 - \frac{1}{2}y = \frac{3}{2}$$

Multiply the entire equation by 2 to eliminate fractions:

$$2y^2 - y = 3$$

Subtract 3 from both sides:

$$2y^2 - y - 3 = 0$$

We can factor the quadratic equation. We need two numbers that multiply to $2 \times (-3) = -6$ and add up to $-1$. These numbers are $-3$ and 2.

$$2y^2 - 3y + 2y - 3 = 0$$

Factor by grouping:

$$y(2y - 3) + 1(2y - 3) = 0$$ $$(y + 1)(2y - 3) = 0$$

This gives two possible values for $y$:

$$y + 1 = 0 \implies y = -1$$ $$2y - 3 = 0 \implies 2y = 3 \implies y = \frac{3}{2}$$

Case 1: $y = -1$

$$\log_3 x = -1$$

By the definition of a logarithm ($b^y = x \iff \log_b x = y$), we have:

$$x = 3^{-1}$$ $$x = \frac{1}{3}$$

Case 2: $y = \frac{3}{2}$

$$\log_3 x = \frac{3}{2}$$

By the definition of a logarithm, we have:

$$x = 3^{3/2}$$ $$x = \sqrt{3^3}$$ $$x = \sqrt{27}$$ $$x = \sqrt{9 \times 3}$$ $$x = 3\sqrt{3}$$

Both solutions are valid since $x > 0$ for $\log_3 x$.

The solutions for $x$ are $\frac{1}{3}$ and $3\sqrt{3}$.

$$x = \frac{1}{3}, 3\sqrt{3}$$