Is the information sufficient 2?

By Definition,

$\displaystyle \Gamma\left( n + \cfrac{1}{2} \right) = \cfrac{(2n - 1)!!}{2^n} \times \sqrt{\pi}$ .

manipulating the equation in the problem above yields

$\displaystyle \begin{aligned} \Gamma\left( n + \cfrac{1}{2} \right) & = \Gamma(x) = \cfrac{105}{16} \times \sqrt{\pi} \\ & = \cfrac{105}{2^4} \times \sqrt{\pi} = \cfrac{\big( 2(4) - 1 \big)!!}{2^4} \times \sqrt{\pi} = \cfrac{(2n - 1)!!}{2^n} \times \sqrt{\pi} \\ n & = 4 \\ & \boxed{x = 4 + \cfrac{1}{2} = \cfrac{9}{2} = 4.5} \end{aligned}$ .

$\begin{aligned} \frac {105\sqrt 2}{16} & = \sqrt{\frac 2\pi} \Gamma (x) \\ \implies \Gamma (x) & = \frac {105\sqrt \pi}{16} = \frac {7!!\sqrt \pi}{2^4} = \Gamma \left(\frac 92\right) & \small \color{#3D99F6} \text{See note.} \\ \implies x & = \frac 92 = \boxed{4.5} \end{aligned}$

Note:

Consider the following identity:

$\small \begin{aligned} \Gamma(s+1) & = s\Gamma(s) \\ \implies \Gamma \left(\frac 92 \right) & = \frac 72 \Gamma \left(\frac 72 \right) = \frac 52 \cdot \frac 72 \Gamma \left(\frac 52 \right) = \frac 32 \cdot \frac 52 \cdot \frac 72 \Gamma \left(\frac 32 \right) = \frac 12 \cdot \frac 32 \cdot \frac 52 \cdot \frac 72 \Gamma \left(\frac 12 \right) = \frac {7!!\sqrt \pi}{2^4} \end{aligned}$

Note that $\small \Gamma \left(\dfrac 12 \right) = \sqrt \pi$