如图所示为一个三角形 $PQR$。 点 $M$ 在 $QR$ 上，使得 $QM = 1$ 个单位，且 $MR = 3$ 个单位。 角 $PRQ = 90^\circ$，且 $PR$ 为 $x$ 个单位。角 $QPR$ 和 $MPR$ 分别记作 $\alpha$ 和 $\beta$，如图所示，其中 $0^\circ < \alpha < 45^\circ$。 已知 $\tan(\alpha - \beta) = \frac{1}{8}$，求 $x$ 的值。

1. Express $\tan \alpha$ and $\tan \beta$ in terms of $x$ using the right-angled triangles.

In $\triangle PQR$, $QR = QM + MR = 1 + 3 = 4$.
$\tan \alpha = \frac{QR}{PR} = \frac{4}{x}$

In $\triangle PMR$, $MR = 3$.
$\tan \beta = \frac{MR}{PR} = \frac{3}{x}$

2. Apply the compound angle formula for $\tan(\alpha - \beta)$.

$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

3. Substitute the expressions into the formula and simplify to form an equation in $x$.

$\frac{1}{8} = \frac{\frac{4}{x} - \frac{3}{x}}{1 + \left(\frac{4}{x}\right)\left(\frac{3}{x}\right)}$
$\frac{1}{8} = \frac{\frac{1}{x}}{1 + \frac{12}{x^2}}$
$\frac{1}{8} = \frac{x}{x^2 + 12}$

4. Solve the resulting quadratic equation.

$x^2 + 12 = 8x$
$x^2 - 8x + 12 = 0$
$(x - 2)(x - 6) = 0$
$x = 2 \text{ or } x = 6$

5. Reject the invalid root using the given condition for $\alpha$.

Given $0^\circ < \alpha < 45^\circ$, we must have $\tan \alpha < \tan 45^\circ$, so $\tan \alpha < 1$.
If $x = 2$, $\tan \alpha = \frac{4}{2} = 2 > 1$ (rejected).
If $x = 6$, $\tan \alpha = \frac{4}{6} = \frac{2}{3} < 1$ (accepted).
Therefore, $x = 6$.