Determine the units of the specific rate constant k given the rate law d[x]/dt = k[B]^(3/4).

The rate equation is given as:

$$\frac{d[x]}{dt} = k[B]^{3/4}$$

The term $\frac{d[x]}{dt}$ represents the rate of reaction. Since $[x]$ is concentration in M (Molarity) and $t$ is time in seconds (s), the units of the rate are $\frac{\text{M}}{\text{s}}$.
The term $[B]$ represents concentration, so its unit is M. Therefore, the unit of $[B]^{3/4}$ is $\text{M}^{3/4}$.

From the equation $\frac{d[x]}{dt} = k[B]^{3/4}$, we can express $k$ as:

$$k = \frac{\frac{d[x]}{dt}}{[B]^{3/4}}$$

$$\text{Units of } k = \text{M}^{1} \cdot \text{s}^{-1} \cdot \text{M}^{-3/4}$$

Combine the terms with M:

$$\text{Units of } k = \text{M}^{(1 - 3/4)} \cdot \text{s}^{-1}$$ $$\text{Units of } k = \text{M}^{(4/4 - 3/4)} \cdot \text{s}^{-1}$$ $$\text{Units of } k = \text{M}^{1/4} \cdot \text{s}^{-1}$$

The units of the specific rate constant $k$ are $\boxed{\text{M}^{1/4}\text{s}^{-1}}$.