Logarithmic inequality (looking for a better solution) $1+\sqrt{17-\log_{x}{2}} \cdot \log_{2}{x^7} \geq \log_{2}{x^{27}}$
Let $t = \log_{2}{x}$. Then we get (taking account of the fact that $x>0$ and $x \ne 1$
$$1+\sqrt{17-\frac{1}{t}}\cdot 7t \geq 27t$$ or
$$\sqrt{17-\frac{1}{t}} \cdot 7t \geq 27t-1 \tag{1}$$
It's clear that $17-\frac{1}{t} \geq 0$ or else the root is not defined so either $t<0$ or $t\geq \frac{1}{17}$. 
Then I consider two cases: when $t<0$ and we can divide by $t$ and reverse the inequality or when $t\geq \frac{1}{17}$ and we can divide by $t$ safely. This way, I am able to get the correct answer. 
Do you think this is the most rational approach to the problem? 
At first, I mistakenly simplified the inequality by dividing the left part by $\sqrt{t}$ but this is obviously not correct for all values of $t$. 

As you wrote, we have to have
$$t< 0\qquad\text{or}\qquad t\ge\frac{1}{17}\tag1$$
Now, we have
$$\begin{align}&1+\sqrt{17-\frac{1}{t}}\cdot 7t \geq 27t
\\\\&\iff 7t\sqrt{17-\frac{1}{t}} \geq 17t-1+10t
\\\\&\iff 7t\sqrt{17-\frac{1}{t}}\geq t\left(17-\frac 1t\right)+10t
\\\\&\iff t\left(17-\frac 1t\right)-7t\sqrt{17-\frac 1t}+10t\le 0
\\\\&\iff t\left(\left(17-\frac 1t\right)-7\sqrt{17-\frac{1}{t}}+10\right)\le 0
\\\\&\iff t\left(\sqrt{17-\frac{1}{t}}-2\right)\left(\sqrt{17-\frac{1}{t}}-5\right)\le 0\tag2\end{align}$$
Now, let us consider $(1)$.
When $t< 0$, we have
$$\begin{align}(2)&\iff \left(\sqrt{17-\frac{1}{t}}-2\right)\left(\sqrt{17-\frac{1}{t}}-5\right)\color{red}{\ge} 0
\\\\&\iff \sqrt{17-\frac{1}{t}}\le 2\qquad\text{or}\qquad \sqrt{17-\frac{1}{t}}\ge 5
\\\\&\iff 17-\frac 1t\le 4\qquad\text{or}\qquad 17-\frac 1t\ge 25
\\\\&\iff 17t-1\ge 4t\qquad\text{or}\qquad 17t-1\le 25t
\\\\&\iff t\ge -\frac 18\end{align}$$
When $t\ge\frac{1}{17}$, we have
$$\begin{align}(2)&\iff \left(\sqrt{17-\frac{1}{t}}-2\right)\left(\sqrt{17-\frac{1}{t}}-5\right)\le 0
\\\\&\iff 2\le \sqrt{17-\frac{1}{t}}\le 5
\\\\&\iff 4\le 17-\frac 1t\le 25
\\\\&\iff 4t\le 17t-1\le 25t
\\\\&\iff t\ge\frac{1}{13}\end{align}$$
So, we get
$$-\frac 18\le t <0\qquad\text{or}\qquad t\ge\frac{1}{13}$$
Hence, the answer is
$$\color{red}{2^{-1/8}\le x< 1\qquad\text{or}\qquad x\ge 2^{1/13}}$$
