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My buddies little brother brought me this question he was assigned for homework. Either I'm crazy, the teacher messed up, or it's a trick question.

I'm assuming $x\in\mathbb R$.

Solve for $x$

$$\log_2(x-2) - \log_2(x+2) = 2$$ Since $\log_2(x-2)$ is only defined for $x >2 $, and since $\log_2(x+2) > \log_2(x-2)$ over [2, $\infty)$, there exists no $x$ that satisfies the equation?

Right??

robjohn
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Colin Martell
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    Actually, $\log_2(x-2)$ is defined only for $x>2$, but otherwise what you say is correct. Perhaps there’s a typo, and it was supposed to be $$\log_2(x+2)-\log_2(x-2)=2;.$$ – Brian M. Scott Aug 21 '13 at 06:31
  • @BrianM.Scott Ah, yes. x>2 thank you. OK at least I know I'm sane. I've been staring that thing for a while trying to figure out where I went wrong. – Colin Martell Aug 21 '13 at 06:33

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Good intuitive approach! Alternately, we can use log rules to see that, if $x$ is a solution to the equation, then $$\log_2\frac{x-2}{x+2}=2,$$ meaning that $$\frac{x-2}{x+2}=2^2=4,$$ so that $$x-2=4(x+2)=4x+8,$$ which we can solve to determine that $x=-\frac{10}3.$ But then neither $x+2$ nor $x-2$ is positive, so this is certainly not allowed in the original equation. Thus, there are no real solutions.

Cameron Buie
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Actually, $\log_2(x-2)$ is defined only for $x>2$, but otherwise what you say is correct. Perhaps there’s a typo, and it was supposed to be $$\log_2(x+2)-\log_2(x-2)=2\;.$$ That leads fairly straightforwardly to the solution $x=\frac{10}3$, using methods appropriate to the level.

Brian M. Scott
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If you just want to solve it

$\log_2(x-2)-\log_2(x+2)=2$ or $\log_2(\frac{x-2}{x+2})=2$

which is equivalent to $\frac{x-2}{x+2}=4$ which gives $x=\frac{-10}{3}$.

robjohn
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Shobhit
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$\log_2{\frac{x-2}{x+2}}=\log_2{4}$

$x-2=4(x+2)$

$3x = -10$

$x = -\frac{10}{3}$

Hence equation has no solution in $\mathbb{R}$, since both $\log_2{(x+2)}$ and $\log_2{(x-2)}$ are in $\mathbb{C}$.

baronbrixius
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In some mathematics texts, all log(x) is automatically considered as log|x|.

Thus, if x was found to be -10/3, the original equation still makes sense. Because ....

$log_2(x-2) - log_2(x+2)$

$= log_2|x-2| - log_2|x+2|$

= …

$= log_2|-16/3| - log_2|-4/3|$

$= log_2(16/3) - log_2(4/3)$

$= log_2(4)$

= 2

= RHS

Mick
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    I have never seen this convention personally - but even if I had I would say do not use it because when you use complex numbers it makes lots of sense to have $\log (-1)= \pi i$ and none to have is as 0. – not all wrong Aug 21 '13 at 18:02
  • @Sharkos, All ordinary logarithmic functions are well defined only on a restricted domain of (0, +infinity). Thus, log(x) must be regarded as taking the log of a positive number. Instead of saving the trouble of writing log|x| each time, we simply write log(x), or even as log x. For this 'tricky question', log (-1) = log|-1| = log (1), unless specified otherwise. – Mick Aug 22 '13 at 15:13
  • I have no idea whatsoever what "ordinary" is supposed to mean there. For mathematicians, my convention is pretty ordinary, and yours looks weird. Of course my definition requires complex numbers, but it is bad to have two competing definitions and mine makes sense in the long run! In particular, any definition of logarithm should satisfy $\exp(\log c)= c$ everywhere the logarithm is defined or mathematicians will start to cry. – not all wrong Aug 22 '13 at 18:59
  • @Sharkos, “ordinary” = “common” and has no special meaning. I just don’t want to mix up with the term “common log”. That is all.

    Many books adopted the “agreed” abbreviation for simplicity:-

    1. Simply stating the law as ln (u/v) = ln u – ln v, rather than formally as ln |u/v| = ln |u| - ln |v|.

    2. Chrystal’s ‘Textbook of Algebra (V.II-293)’ says “…, then the [LHS]of [Log w = log |w| + i amp(w)] will vary continuous through all values which Log w can assume for a given value of |w|”.

    3. Gerrish’s ‘Pure Mathematics I (P114)’ says “We usually omit the modulus in Answers to Exercises.”

     – Mick Aug 23 '13 at 03:46
  • up to multiples of $2\pi i$ this always holds for my definition. 2) that definition defines my complex logarithm using the value of the logarithm for positive numbers; it doesn't support your case since $\log(-1)$ is never used whilst $\mathrm{Log}(-1)= 0+ i\cdot\pi$ just as I said above. 3) That sounds like sloppy laziness, though without context it's difficult to judge, since there is no good reason for defining logs of negatives like this of which I'm aware. The only thing I can think of is $\int 1/x$; but being lazy is bad in mathematics so this should not be done.
    • – not all wrong Aug 23 '13 at 07:30
  • Also "common" definitions of log in the mathematical community are exactly of the $\mathrm{Log}$ type as I mentioned before. Professional mathematicians think $\log (-1)$ is $\pi i$ or a shifted version thereof; that is common there. If you've seen the other convention more commonly, then I recommend starting to use modulus signs. It's better in the long run. – not all wrong Aug 23 '13 at 07:34