Have three questions as stated below, that concern respectively with the interpretation of the result obtained (with some proof, if possible), and why my solution set is different than the book's .:

The different absolute value terms are stated below with the $4$ intervals defined due to them :
(i) $x+2 : x+2\lt 0, x+2\ge 0\implies x\lt -2, x\ge -2$
(ii) $2^{x+1}-1 : 2.2^x+1\lt 0, 2.2^x+1\ge 0\implies 2^x\lt -\frac12, 2^x\ge -\frac12\implies 2^x\lt 2^{-1}, 2^x\ge 2^{-1}\implies x\lt -1, x\ge -1$
The second case can be alternatively solved by :
$2^{x+1}-2^0 : x+1\lt 0, x+1\ge 0\implies x\lt -1, x\ge -1$

So, the $2$ points on the real line are : $x=-2, -1$, leading to $3$ intervals out of $4$:
(i) $(-\infty,-2)$, (ii) $[-2,-1)$, (iii) $[-1,\infty)$

The $3$ intervals with their analysis is given below:
(a) $x\lt -2$ : $2^{-(x+2)} -2^{x+1}+1 = 2^{x+1}+1\implies 1 = 2^{2x+4}\implies 2x+4 =0\implies x = -2$. The value obtained lies outside the stated interval.
(b) $-2\le x\lt -1$ : $2^{x+2} +2^{x+1}-1 = 2^{x+1}+1\implies x+2 = 1\implies x =-1$.
The value obtained lies outside the stated interval.
(c) $x\ge -1$ : $2^{x+2} -2^{x+1}+1 = 2^{x+1}+1\implies x+2 = 2x+2\implies x =0$
The value obtained lies in the stated interval.

So, only the 3rd case satisfies the given equality.

Let us try by substituting values for different cases as below:

(1) : For $x=0$ substituting back gives: $4 -2+1 = 3$.
Trying other values as below :
$x=-1$ yields: $2=2$ which is also correct.
$x=1$ yields: $8 - 3 = 5$.

So, all $3$ values satisfy in the third case. But, lack a proof that if get a value inside the interval, then all values satisfy it.

Question 1. Can I interpret it to mean that getting:
(i) the equality $0=0$,
(ii) or a value of $x$ in the stated interval (as above got $x =0$)
shows a tautology, i.e. all values in the interval satisfy it.

(2) : Similarly, for second interval trying the value $x=-\frac32$ by substituting back, get:
$\sqrt{2} -\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}+1\implies 2-1 = 1+\sqrt{2}$, which is false.

(3) : Similarly, for the first interval trying the value $x=-2$ by substituting back, get:
$1 -\frac12= -\frac12$, which is tautology, but then where I erred above is not clear.
I try out some values as below to check:
(a) $x=-2$: $1 -(-\frac12)= \frac32$, which is true.

Question 2: Why my above analysis for first case is proved wrong. In fact, the value $x=-2$ makes my analysis for the second interval also wrong.

Question 3 : My third question is about the solution given that differs from mine, as shown below in the page #40 of the book 'Elementary Mathematics, by - Dorofeev, Potapov, Rozov'. I request how it is possible that both are correct; else which one is wrong.
My solution is not having the point $x=-3$.

enter image description here


If $x<-2$, the equation becomes






If $x \ge -1$,


$$2^{x+2}=2\cdot 2^{x+1}=2^{x+2}$$

Hence we have $$x+2=x+2$$ (rather than $x+2=2x+2$)

Hence the equality is satisfied by every $x$ that satisfies $x \ge -1$.


If $x=-2$,




  • $\begingroup$ Thanks a lot. Request help in Question #2, where $x=-2$ is satisfying the equality; as shown in the OP, and below too: $2^0 +\frac12 = 2^0+ \frac 12$. I mean that there should be no solution in the second case, and $x=-2$ lies in that interval. $\endgroup$ – jiten Jun 4 '18 at 1:04
  • 1
    $\begingroup$ $x=-2$ is not an answer, I think you made a sign error. $\endgroup$ – Siong Thye Goh Jun 4 '18 at 1:09
  • $\begingroup$ Yes, I ignored absolute value sign. Thanks a lot. This question was a very old sticking point for me. $\endgroup$ – jiten Jun 4 '18 at 1:11
  • $\begingroup$ Have problem solution given at: math.stackexchange.com/a/460988/424260, which am unable to understand. This problem's graphical soln. takes $|x-a|\lt x^2-3$, & has right arm of the graph for $|x-a|$ first touching parabola for negative $a$, & then the left arm for positive $a$, with range between these two values of $a$ being asked for. Am unable to get the stated values as $x^2+x-3\lt a\lt -x^2+x+3$ has both l.h.s & r.h.s with roots given by $x= \frac{-1\pm \sqrt{13}}{2}$. So, the second para. is incomprehensible to me. Could have placed separate post, but this duplicate exists. $\endgroup$ – jiten Jun 4 '18 at 7:19
  • $\begingroup$ Please tell me if need place a separate post for the above comment's contents. $\endgroup$ – jiten Jun 4 '18 at 7:57

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