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Solve the following logarithmic inequality: $\frac{2\log_a x}{1+2 \log_a x} < \log_2^2 x$ assuming that $a \in (0,1) \cup (1,+\infty)$ and $\lvert 2\log_a x \rvert < 1$

The answer should look like this:

for $a \in (0,1)$ : $x \in \left(1, \frac{1}{\sqrt{a}}\right)$

for $a > 1$ : $x \in \left(\frac{1}{\sqrt{a}}, 1\right)$

For well-defined $(a,\ x)$ and $\begin{cases}\log_ax=t\\\log_a2=p\end{cases}$ the given inequality is equivalent. $$\frac{2t}{1+2t}<\left(\frac{t}{p}\right)^2$$ $$\frac{2p^2t}{1+2t}-\frac{t^2(1+2t)}{1+2t}<0$$ $$-t(2t^2+t-2p^2)(1+2t)<0$$ $$-4t\left(t-\frac{-1-\sqrt{1+16p^2}}{4}\right)\left(t-\frac{-1+\sqrt{1+16p^2}}{4}\right)\left(t+\frac{1}{2}\right)<0$$

$$t_2=\frac{-1-\sqrt{1+16p^2}}{4}<t_4=-{1\over2}<t_1=0<t_3=\frac{-1+\sqrt{1+16p^2}}{4}$$ Thus, the inequality is satisfied for: $$t\in(-\infty;t_2)\cup(t_4;t_1)\cup(t_3;+\infty)$$

What to do next to get this result, or maybe you have some other way?

Fty56
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  • I think you could try logarithm base change, so you can work with the same base and $t=\log_2 x$ is the only variable you need instead of working with two. – Turquoise Tilt Sep 19 '23 at 10:10

1 Answers1

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In general, let's consider $t_1 < \log_a x < t_2$. Then, we need to look at two cases:

  1. $a>1$. In this case $\log_a$ is strictly increasing function with strictly increasing inverse $x\mapsto a^x$, so $$t_1 < \log_a x < t_2\iff a^{t_1} < x <a^{t_2}.$$ Note that if $t_1 = -\infty$ or $t_2 = \infty$, the conclusion still works, but we think of $a^{t_1}$ (or $a^{t_2}$) as limits, i.e. $a^{-\infty} = 0$ and $a^{\infty} = \infty$.

  2. $0<a<1$. In this case $\log_a$ is strictly decreasing function with strictly decreasing inverse $x\mapsto a^x$, so $$t_1 < \log_a x < t_2\iff a^{t_1} > x > a^{t_2}.$$ Note that if $t_1 = -\infty$ or $t_2 = \infty$, the conclusion still works, but we think of $a^{t_1}$ (or $a^{t_2}$) as limits, i.e. $a^{-\infty} = \infty$ and $a^{\infty} = 0$.

What you did is correct, but you forgot the condition that $|2\log_ax|< 1$, i.e. $|2t| < 1$, i.e. $-1/2< t < 1/2$. This eliminates $t\in (-\infty, t_2)$ and you need to check whether $t_3 \leq 1/2$ or not, to decide what to do with interval $(t_3,+\infty)$.

Ennar
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  • So is this the final answer yet? – Fty56 Sep 19 '23 at 10:28
  • @Fty56 if by the final answer you mean to find for which $x$ does the inequality hold, then yes, this is how you reach it from the work you've done yourself. – Ennar Sep 19 '23 at 10:30
  • But such a number is not included in the solution- $a^{\frac{-1+\sqrt{1+16\log_a^22}}{4}}$ – Fty56 Sep 19 '23 at 10:37
  • @Fty56, I don't understand what you mean. What I wrote is generally true for logarithms and exponential functions. – Ennar Sep 19 '23 at 10:39
  • And how to determine the order between $t_3$ and $1\over2$:

    $t_3\le{1\over2}\iff |p|\le{\sqrt2\over2}\iff a>1$

     – Fty56 Sep 19 '23 at 10:48
  • @Fty56 substitute back $p = \log_a2$ and use what I wrote in my answer. If you don't want to consider cases whether $a$ is greater or less than $1$, you can use $1/p = \log_2a$ – Ennar Sep 19 '23 at 11:11
  • After all, I used this substitution? – Fty56 Sep 19 '23 at 11:14
  • @Fty56 I feel there must be a language barrier of sorts going on here. Substitute back means that instead of $|p|\leq \sqrt 2/2$ you write $|\log_a2| \leq \sqrt 2/2$. – Ennar Sep 19 '23 at 11:16
  • $0 < a \leq \frac{1}{\sqrt{2}} \quad \text{or} \quad a \geq \sqrt{2}$ – Fty56 Sep 19 '23 at 11:27
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    @Fty56, you should get $0<a\leq 2^{-\sqrt{2}}\lor a\geq 2^{\sqrt{2}}$. – Ennar Sep 19 '23 at 11:31
  • Actually my mistake. So what to do next about it? – Fty56 Sep 19 '23 at 12:26
  • @Fty56 split it into cases $a\in (0, 2^{-\sqrt 2}]$, $ a\in (2^{-\sqrt 2},1)$, $a\in (1, 2^{\sqrt 2})$, $a\in [2^{-\sqrt 2},\infty)$. For some cases you'll have $t\in (t_4,t_1)$ and for some you'll have $t\in(t_4,t_1)\cup (t_3,1/2)$. Then, substitute back $ t= \log_ax$ and solve the inequalities according to what I wrote. – Ennar Sep 20 '23 at 08:40