Confusion regarding calculating the pH of a salt of weak acid and weak base
Another approach, I believe, is this:

Definitions
$$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right) \\ 2\,\mathrm{pH}=\log(\frac{K_b}{K_a\,K_w})$$

It suggests equations' sum, like this:  

$$\ce{HA &lt;=&gt; A- + H+} \tag{$\ce{K_a=\frac{[A-][H+]}{[HA]}}$}$$ 
$$\ce{BOH &lt;=&gt; B+ + HO-} \tag{$\ce{K_b=\frac{[HO-][B+]}{[BOH]}}$}$$ 
$$\ce{A- + B+ + H2O &lt;=&gt; AH + BOH}\tag{$\ce{K=\frac{[AH]\,[BOH]}{[A-][B+]}=\frac{K_w}{K_a\,K_b}}$}$$
Finding $\ce{[H+]}$ in relation to $\ce{K}$

If salt's concentration is $\ce{C_s}$, and the mass balance:

$$\ce{C_s = [AH] + [A-] = [BOH] + [B+] } \tag{Mass balance}$$ 

Dividing by $\ce{C_s}$ we define $\alpha$ and $\beta$ such that
$$\ce{\frac{C_s}{C_s} = \alpha{} + (1-\alpha) = \beta{} + (1-\beta)} \tag{Mass balance}$$ 

And if we admit that $\alpha$, $\beta &lt;&lt;1$ which also implies $\alpha\approx\beta$ (this is possibly a weakness of the answer) then $$\ce{\alpha^2\approx\frac{K_w}{K_a\,K_b}}$$

And $\ce{K_a=\frac{(1-\alpha)[H+]}{\alpha}\approx\frac{[H+]}{\alpha}}$

$$(\frac{[H^+]}{K_a})^{2}\approx \frac{K_w}{K_a\,K_b}$$
$$[H^+]^{2}\approx \frac{K_w\,K_a}{K_b}$$

And we get the equation.