The problems in my sophomore workbook are marked with three colors that signify how hard is the marked problem: green for D and C, yellow for B and A and red for advanced students. All problems are also ordered from the easiest to the hardest.

This problem is marked red and appears last in the section of logarithmic problems, which means that it's the hardest logarithmic problem in the whole workbook. I know what its solution is (there are solutions at the end of the workbook), but I'd like to see how do we arrive at it.

I wrote the solution here, but it's hidden until it's hovered over. I don't want to spoil fun to those who want to find it by themselves.

Find the value(s) of $x$ when the left-hand side is equal to the right-hand side.


The solution is




Since $$\sqrt{\frac{1+x^2}{2x}\pm1}= \sqrt{(x\pm1)^2\over 2x} = {|x\pm1|\over \sqrt{2x}}$$ we have $$\frac{\sqrt{\frac{1+x^2}{2x}+1}-\sqrt{\frac{1+x^2}{2x}-1}}{\sqrt{\frac{1+x^2}{2x}+1}+\sqrt{\frac{1+x^2}{2x}-1}}={|x+1|-|x-1|\over |x+1|+|x-1|}$$

Let $f(x)=|x+2|+|x-2| $, then $f(x)=4$ for $x\in[-2,2]$ and for $|x|>2$ we have $f(x)>4$.

So, for $|x|>2$ we have $\log_2f(x)-11/9>7/9$ so we have $|x+1|>8|x-1|$

a) if $x>2$ we get $x<9/7$ a contradiction and

b) if $x<-2$ we get $x>7/9$ a contradiction again.

So we are left with the $|x|\leq 2$ so we have $$|x+1|=8|x-1|$$

Can you finish?

  • $\begingroup$ I'll give it a go. $\endgroup$ – Hanlon May 27 '18 at 9:54
  • $\begingroup$ To hard? @Hanlon $\endgroup$ – Aqua May 27 '18 at 10:11

Simplify the LHS: $$\frac{\sqrt{\frac{1+x^2}{2x}+1}-\sqrt{\frac{1+x^2}{2x}-1}}{\sqrt{\frac{1+x^2}{2x}+1}+\sqrt{\frac{1+x^2}{2x}-1}}=\frac{\sqrt{\frac{(x+1)^2}{2x}}-\sqrt{\frac{(x-1)^2}{2x}}}{\sqrt{\frac{(x+1)^2}{2x}}+\sqrt{\frac{(x-1)^2}{2x}}}\overbrace{=}^{x>0}{|x+1|-|x-1|\over |x+1|+|x-1|}.$$ Hence: $${|x+1|-|x-1|\over |x+1|+|x-1|}=\log_2(|x-2|+|x+2|)-\frac{11}{9}.$$ Consider the intervals: $$x\in (0,1]: \qquad {2x\over 2}=\log_2 4-\frac{11}{9} \Rightarrow x=\frac79;\\ x\in (1,2]: \qquad {2\over 2x}=\log_2 4-\frac{11}{9} \Rightarrow x=\frac97;\\ x\in (2,+\infty): \qquad {2\over 2x}=\log_2 (2x)-\frac{11}{9} \Rightarrow \log_2x=\frac 1x+\frac29 \Rightarrow \emptyset, \text{because}:\\ \log_2x\ge 1>\frac{13}{18}\ge \frac 1x+\frac29, x\ge 2.$$


Hint: multiplying the left-hand side's numerator and denominator by the numerator obtains $r-\sqrt{r^2-1}$, and this is just $\exp -|\ln x|$. So we want to solve $\exp -|\ln x|+\frac{11}{9}=\log_2 (|x-2|+|x+2|)$.

  • $\begingroup$ This is not true, put $x= 2$ and you will see. $\endgroup$ – Aqua May 27 '18 at 8:10
  • $\begingroup$ @ChristianF Sorry; fixed. $\endgroup$ – J.G. May 27 '18 at 8:21
  • $\begingroup$ This is stil not true. try $x=1/2 $\endgroup$ – Aqua May 27 '18 at 8:25
  • $\begingroup$ @ChristianF Sorry; I had to go offline for a few hours. $\endgroup$ – J.G. May 27 '18 at 13:53

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