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I'm trying to solve the following inequality $\dfrac{(\log_2 (8x) \times \log_{x/8} 2)}{\log_{x/2} 16} \leq 0.25$

Wolfram alpha gives the answer $(0, 0.5], [1,8)$ but surely $x \not= 2$ since log to base $1$ is undefined. But is the fact that it basically shrinks the fraction down to $0$ sufficient to satisfy this inequality? Could someone clear this up for me?

Wolfram link is here.

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You can otherwise write the inequality as: $$\frac{\log_2 8x \times \log_{\frac{x}{8}}2 \times \log \frac{x}{2}}{\log 16} \leq 0.25$$ $$\Rightarrow \frac{\log \frac{x}{2} \times \log 8x}{\log 16 \times \log \frac{x}{8}} \leq 0.25$$ What we just did is to use the logarithmic identity: $\log_{a}b = \frac{\log b}{\log a}$. Now we can see that $x$ can take the value of $2$ because due to this identity, we now have the numerator as $0$ which is less than $0.25$.

Hope it is much clearer to you now.

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  • $\begingroup$ I'm still a little conflicted, on the one hand I feel like it makes sense but another part of me questions it $\endgroup$ – user142702 Dec 23 '16 at 13:53
  • $\begingroup$ @user142702 Yes, $2$ is very well a solution. Just rearrange the terms and see. $\endgroup$ – Rohan Dec 23 '16 at 13:56
  • $\begingroup$ So am I correct I thinking $log_1 16$ is undefined but $\dfrac{1}{log_1 16}$ is 0 as we can flip the fraction? $\endgroup$ – user142702 Dec 23 '16 at 14:08
  • $\begingroup$ Yes, you are absolutely correct. $\endgroup$ – Rohan Dec 23 '16 at 14:09
  • $\begingroup$ And this thus justifies why $x=8$ is not a solution because $log_1 2$ is undefined and on the numerator $\endgroup$ – user142702 Dec 23 '16 at 14:13
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Put everything on the base $2$:

$$\dfrac{(\log_2 (8x) \times \log_{x/8} 2)}{\log_{x/2} 16} \leq 0.25$$

$$\dfrac{\log_2 (8x) }{\log_{x/2} 16}\times \frac{\log_{2} 2}{\log_{2} (x/8)} \leq 0.25$$

$$\log_2(8x)\times\dfrac{\log_2 (x/2) }{\log_{2} 16}\times \frac{\log_{2} 2}{\log_{2} (x/8)} \leq 0.25$$

$$\log_2(8x)\times\dfrac{\log_2 (x/2) }{4}\times \frac{1}{\log_{2} (x/8)} \leq 0.25$$

$$(3+\log_2 x)\cdot (\log_2 x -1)\cdot \frac{1}{\log_{2} x -3} \leq 1$$

write $\log_{2} x=y$ and then

$$\frac{(y+3)(y-1)}{y-3}\le 1 \Rightarrow \frac{y^2+y}{y-3} \le 0$$

So, $y \le -1$ or $0 \le y <3$.

$$y \le -1 \Rightarrow \log_2 x \le -1 \Rightarrow x \le 1/2 \quad (1)$$

$$0 \le y <3 \Rightarrow 0 \le \log_2 x <3 \Rightarrow 1 \le x < 8 \quad (2)$$

But now we have to check the boundary for the problem. Which are:

$$8x >0 \Rightarrow x>0 \quad (3)$$

$$x/8 >0 \Rightarrow x>0 \quad x/8 \ne 1 \Rightarrow x \ne 8 \quad (4)$$

$$x/2 >0 \Rightarrow x>0 \quad x/2 \ne 1 \Rightarrow x \ne 2 \quad (5)$$

Making the intersection of $(1),(2),(3),(4),(5)$ we get:

$$0<x \le 1/2 \quad \text{or} \quad 1 \le x <8 \quad \text{and} \quad x \ne 2$$

or writing in another way:

$$0<x \le 1/2 \quad \text{or} \quad 1 \le x <2 \quad \text{or} \quad 2<x<8$$

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we write your inequality in the form $$\frac{\frac{\ln(8x)}{\ln(x/8)}}{\frac{\ln(16)}{\ln(x/2)}}\le \frac{1}{4}$$ simplifying this we get $$\frac{(3\ln(2)+\ln(x))(\ln(x)-\ln(2))}{4\ln(2)(\ln(x)-3\ln(2))}\le \frac{1}{4}$$ simplifying again we get $$\frac{\ln(x)^2+2\ln(2)\ln(x)-3\ln(2)^2}{(\ln(x)-3\ln(2))\ln(2)}\le 1$$ can you proceed? simplifying again we get $$\frac{\ln(x)(\ln(x)+\ln(2))}{\ln(x)-3\ln(2)}\le 0$$ doing case work we obtain $$0<x\le \frac{1}{2}$$ or $$1\le x< 8$$ and $$x\ne 2$$

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  • $\begingroup$ Hi, I have worked through in a similar way to Arnaldo and obtained $(0, 0.5], [1,2), (2,8)$ but my issue is whether x=2 is a solution or not. Is it a solution? $\endgroup$ – user142702 Dec 23 '16 at 13:51
  • $\begingroup$ $$x=2$$ is not a solution since the loagarithm of the denominator is not defined $\endgroup$ – Dr. Sonnhard Graubner Dec 23 '16 at 14:19
  • $\begingroup$ Could you elaborate further? This is what I thought but having observed Rohan's answer, it seems to be agreeable $\endgroup$ – user142702 Dec 23 '16 at 15:52
  • $\begingroup$ Or I guess as a contrapositive, could you explain why Rohan's explanation doesn't work? $\endgroup$ – user142702 Dec 23 '16 at 15:52

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