Logarithmic inequality (looking for a better solution)

Question

$1+\sqrt{17-\log_{x}{2}} \cdot \log_{2}{x^7} \geq \log_{2}{x^{27}}$

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 $t$.

Answer 1

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 $$\color{red}{2^{-1/8}\le x< 1\qquad\text{or}\qquad x\ge 2^{1/13}}$$

Answer 2

Most rational approach? Well, that depends on what you mean by rational. In any case all approaches are valid as long as they can be proved correct.

As for the problem, another way is to consider cases when $t\ge 0$ or when $t\le 0.$ For this first case, you can square both sides of your equation $(1)$ to get a quadratic in $t,$ which is easily analysed. For the case when $t<0,$ first multiply both sides by $-1,$ then you may square and proceed as before.

Answer 3

When the reality condition demands $17-1/t>0~~~(1)$ and if $t>1/27~~~(2)$ you can square you eqation in $t$ to get $$104t^2+5t-1.0 \Rightarrow (t-1/13)(t+1/8)>0 \Rightarrow t<-1/8~\mbox{or}~t>1/13.$$ Only $t>1/13$ satisfies (2). This gives $x>2^{1/13}$ as the only solution to your log-equation.

For $t<0$ the sign of as $2t-1<0$ we will get $$104t^2+5t-1<0 \Rightarrow ((t-1/13)((t+/8)<0 \Rightarrow $-1/8<t<1/13.$$ Then the solution is $-1/8<t<0.$ Next we have to solve $$\log_{2} x < 0 \Rightarrow \log_2 x< \log_2 1 \Rightarrow x<1$$ and $$t >-1/8 \Rightarrow \log_2 x >-1/8 \Rightarrow \log_2 x >\log_2 2^{-1/8} \Rightarrow x>2^{-1/8}.$$ Finally, $t<0$ case gives $$ 2^{-1/8} < x <1.$$