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What I was able to get was $\log_2(x^2+2x-7) = \frac{1}{2}\log_2{(x-3)^2}$. Wolfram Alpha says that the solution is $-5$ but I can't get it right? Can someone explain it from here? Thanks.

LearningMath
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    Note that $\frac{1}{2} \log(a^2) = \log(,|a|,),$, and $\log$ is injective. Also note that $x=2$ is another solution to the equation you derived, but not to the original one. – dxiv Jun 18 '17 at 02:35

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Hint: $$\log_2(x^2+2x-7) = \frac{1}{2}\log_2{(x-3)^2} $$$$ \log_2(x^2+2x-7) =\log_2{|x-3|} $$$$ x^2+2x-7=|x-3| $$

BAI
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  • I got the same but without the absolute value. What is the absolute value there for? – LearningMath Jun 18 '17 at 02:46
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    $\sqrt{x^2}=|x|$ – Steven Alexis Gregory Jun 18 '17 at 02:47
  • @LearningMath note that $\sqrt (x-3)^2$ must be positive, but it is not necessary for $x-3$. – BAI Jun 18 '17 at 02:52
  • What is the use of the square function here? Shouldn't we just use the power logarithm law? – LearningMath Jun 18 '17 at 02:53
  • @LearningMath it is exactly the power logarithm law----$\frac12\log(x-3)^2=\log((x-3)^2)^{1/2}=\log\sqrt{(x-3)^2}$ – BAI Jun 18 '17 at 03:05
  • Why can't I just move the two in the exponent and multiply it with $\frac{1}{2}$? Thanks. – LearningMath Jun 18 '17 at 11:55
  • @LearningMath do you mean taking exponent on both sides? But you'll still have the same, as $e^{\frac{1}{2}\log_2{(x-3)^2}}=(e^{\log_2{(x-3)^2}})^\frac{1}{2}$ – BAI Jun 18 '17 at 11:59
  • I was referring to the power logarithm law $\log_2{x^n} = n\log_2x$. I just can't see what is the need of the absolute value there? Can you explain it further? Thanks a lot. – LearningMath Jun 18 '17 at 12:24
  • @LearningMath well, it is just the case that $\sqrt{(x-3)^2}\neq x-3$ if we consider all real $x$ (as when $x\lt 3$, $\sqrt{(x-3)^2}=3-x$), so we must need to put on a absolute value. Does it make things clearer? – BAI Jun 18 '17 at 12:29
  • I understand that. I'm asking about this: $\frac{1}{2}\log_2{(x-3)^2} = 2 \times \frac{1}{2} \log_2(x-3) = \log_2{(x-3)}$. Then this: $x^2 +2x-7=x-3$. Why is this wrong? Why should I use the absolute value here? Thanks again. – LearningMath Jun 18 '17 at 12:35
  • @LearningMath I see. But actually $\log$ is just defined on $\mathbb R_+$, you can't take the $\log$ of a negative number without extend the codomain to $\mathbb C$ (and it is not the case, I think, in your question). Thus, it is undefined if $x\lt 3$, and so we must put an absolute value. – BAI Jun 18 '17 at 12:45
  • Then why is it not the case with $x^2+2x-7$? Why is that not in absolute value? – LearningMath Jun 18 '17 at 13:25
  • @LearningMath well, it might be personal habit, that I always reject the inappropriate solution (as when the quadratic turns negative by the solution got) after I found them. But in this case, for the $x-3$ stuff, it is straight forward that both cases should be considered at this moment, as it is also possible for $x\lt 3$ in this question. But for the 'both cases' considering the quadratic, you could ignore it at the time as for the case it turns negative, he solution would be rejected. – BAI Jun 18 '17 at 13:46
  • @LearningMath So two main reasons: 1. Both $x\lt 3$ or $x\ge 3$ are possible for our solution to be in the range, but it is impossible that one solution would turn out that the quadratic becomes negative. (As the RHS is real). For an solution, say $x$, it is possible that $x-3$ is negative or positive, but it is not that the quadratic is negative. 2. $(x-3)^2=(3-x)^2$, and that is the reason we need to consider both cases, in which we express it in the form of a taking of absolute value. – BAI Jun 18 '17 at 13:46
  • @LearningMath so the absolute value is a way for us to express that both cases could exist. I'm sorry that I'm not a native speaker in English and my expression might not deliver the clear meaning. Hope you would understand. – BAI Jun 18 '17 at 13:48
  • @BAI Thanks a lot for the explanation. So, as far as I understand, in the general case, one should always write $x$ in $\log{x}$ as $\left |x \right |$ just to ensure that no solutions are missed, right? Next, why if we use the power law, we actually miss the solution (because we don't consider the case $3-x$)? Does this mean that there is just an implication between $\log_2{(x-3)^2}$ and $2\log_2{(x-3)}$ i.e the two expressions are not equivalent? Also, I didn't quite understand the comment about rejecting the inappropriate solution of the quadratic. Could you explain it again? Thanks a lot. – LearningMath Jun 18 '17 at 19:46
  • @LearningMath yes, the expressions are not equivalent, as $(x-3)^2=(3-x)^2$. And for the rejection, it occurs when we get a solution that contradict to our assumption (as in this case, if the solution makes $x^2-2x-7$ to be negative, then it should be rejected. And also if, in this question, one of the solutions of $x^2+2x-7=x-3$(in which we assumed that $x\ge 3) is smaller than $3$ in the contradict, it should also be rejected) so all in all, we tends to consider more cases as we can see, and reject the ones which is self-contradictory, to ensure that no solution is missed. – BAI Jun 19 '17 at 02:36