Solve the equation ((2)/(3))^x + ((3)/(2))^x = (13)/(6) for x.

Notice that $\left(\frac{3}{2}\right)^x = \left(\left(\frac{2}{3}\right)^{-1}\right)^x = \left(\frac{2}{3}\right)^{-x}$.
Let $y = \left(\frac{2}{3}\right)^x$. Then the equation becomes:

$$y + y^{-1} = \frac{13}{6}$$ $$y + \frac{1}{y} = \frac{13}{6}$$

Multiply the entire equation by $6y$ to eliminate the denominators:

$$6y \cdot y + 6y \cdot \frac{1}{y} = 6y \cdot \frac{13}{6}$$ $$6y^2 + 6 = 13y$$

Rearrange the terms to form a standard quadratic equation:

$$6y^2 - 13y + 6 = 0$$

We can factor the quadratic equation. We need two numbers that multiply to $6 \times 6 = 36$ and add up to $-13$. These numbers are $-4$ and $-9$.

$$6y^2 - 9y - 4y + 6 = 0$$

Factor by grouping:

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

This gives two possible values for $y$:

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

Case 1: $y = \frac{2}{3}$

$$\left(\frac{2}{3}\right)^x = \frac{2}{3}$$

Since $\frac{2}{3} = \left(\frac{2}{3}\right)^1$, we have:

$$\left(\frac{2}{3}\right)^x = \left(\frac{2}{3}\right)^1$$

Therefore, $x = 1$.

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

$$\left(\frac{2}{3}\right)^x = \frac{3}{2}$$

Since $\frac{3}{2}$ is the reciprocal of $\frac{2}{3}$, we can write $\frac{3}{2} = \left(\frac{2}{3}\right)^{-1}$.

$$\left(\frac{2}{3}\right)^x = \left(\frac{2}{3}\right)^{-1}$$

Therefore, $x = -1$.

The values of $x$ are 1 and $-1$.

$$x = 1, -1$$