That's Tricky #1

Note first that $14x - 20 - 2x^{2} = -2(x^{2} - 7x + 10) = -2y.$ So for both $\sqrt{-2y}$ and $\sqrt{y}$ to be real we must have that

$y = x^{2} - 7x + 10 = (x - 2)(x - 5) = 0 \Longrightarrow x = 2$ or $x = 5.$

With $x = 2$ we have that $9\log_{4}(\frac{x}{8}) = 9\log_{4}(\frac{1}{4}) = -9,$ and that $2x - 13 = 4 - 13 = -9.$ Thus $x = 2$ satisfies the given inequality.

With $x = 5$ we have that $9\log_{4}(\frac{x}{8}) = 9\left(\dfrac{\log(5)}{\log(4)} - \log_{4}(8)\right) = 9\left(\dfrac{1 - \log(2)}{2\log(2)} - \log_{4}(4^{\frac{3}{2}})\right) =$

$9\left(\dfrac{1}{2\log{2}} - \dfrac{1}{2} - \dfrac{3}{2}\right) = \dfrac{9}{2\log(2)} - 18,$

where $\log$ is base $10.$ Now $\log(2) = 0.30103...$ is just slightly more than $\dfrac{3}{10},$ and thus

$\dfrac{9}{2\log(2)} - 18 \lt \dfrac{9}{2*\frac{3}{10}} - 18 = 15 - 18 = -3.$

But for $x = 5$ we have that $2x - 13 = 10 - 13 = -3,$ and thus $x = 5$ does not satisfy the given inequality.

(We could have just used a calculator for this last case, but where's the challenge in that. We did use the value for $\log(2),$ but this should be relatively familiar.)

We thus can conclude that $x = \boxed{2}$ is the unique value satisfying the given inequality.