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This came across a discussion amongst Algebra 2 teachers at my school.
We know $a\log x= \log x^a$
Say
$2\log x=5$
$\log x^2 =5$
When $\log x=\log_{10} x$
Solving the first equation yields $x=10^{5/2}$ while the second equation yields $x=\pm\sqrt{10^5}$.
Which solution is correct or does it depend on the equation?

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    $\begingroup$ Question : is $x=-\sqrt{10^5}$ a solution of $2 \log x=5$? $\endgroup$ – Erik Jurriën May 11 '16 at 13:20
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    $\begingroup$ There is no real number $k$ such that $x=10^k$ is negative, so $\log_{10}x$ is undefined for $x$ negative. $\endgroup$ – almagest May 11 '16 at 13:21
  • $\begingroup$ @almagest: You are of course correct in the context of real numbers. One does have, however, that if $x<0$ then each of the values $$\ell_{m,n}=\frac{\ln |x| + (2m+1)\pi i}{\ln 10 + 2n\pi i}$$ is a possible value of $\log_{10} x$ (where "$\ln$" is a real-valued function of a positive real number). The principal value is $$\frac{\ln |x| + \pi i}{\ln 10}$$ which can be thought of as $\log_{10}|x|+i\frac{\pi}{\ln 10}$. $\endgroup$ – MPW May 11 '16 at 13:32
  • $\begingroup$ The answers are not different except for $\pm$, which arises from (as other pointed out) the fact that first equation can't have negative solutions. Otherwise $10^{5/2}=10^{5\cdot{1\over2}}=(10^5)^{1/2}=\sqrt{10^5}$; as you know $a^{1/2}=\sqrt{a}$. $\endgroup$ – Heimdall May 11 '16 at 20:02
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    $\begingroup$ This question is really just, "Why is $x^2 = 4$ not the same as $2x = 4$?" in disguise. (I used a different constant, of course, but the concepts are identical.) $\endgroup$ – jpmc26 May 12 '16 at 8:19
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The equations "$2\log x = 5$" and "$\log x^2=5$" are not equivalent.

The reason is that the first equation implies that $x>0$ while the second does not.

The correct way to move from the first to the second is to conjoin the condition $x>0$. So instead, one can say that "$2\log x = 5$" and "$\log x^2 = 5 \textrm{ where }x>0$" are equivalent.

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    $\begingroup$ This is why it is worthwhile for Algebra II teachers to notice the difference between the statements, "$\log{(x^a)}=a\log{(x)}$" and "$\log{(x^a)}=a\log{(x)}$ when $x\in \mathbb{R}, x>0$." $\endgroup$ – rnrstopstraffic May 11 '16 at 17:31
  • $\begingroup$ @rnrstopstraffic: Yes indeed. Otherwise it's just "symbol pushing" without thought. Admittedly, one often applies such rules in a context where everything is always positive, but it really is worth remembering what assumptions are lurking in the background. +1 for you. $\endgroup$ – MPW May 11 '16 at 17:34
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    $\begingroup$ @rnrstopstraffic: One also needs to qualify $a\in\Bbb R$ (though it may not be clear to Algebra II students why this is a restriction). Note that $\ln(e^{2\pi\mathbf i}) \neq 2\pi\mathbf i\ln(e)$. $\endgroup$ – Marc van Leeuwen May 12 '16 at 8:42
  • $\begingroup$ @MarcvanLeeuwen: And very sadly it is usually not clear to Algebra II teachers as well... $\endgroup$ – user21820 May 12 '16 at 11:09
  • $\begingroup$ @MarcvanLeeuwen: You are right. I didn't include it at that point because Algebra II generally has not introduced the notion of non-real exponents; typically they've barely been introduced to the notion that the exponential function is, in fact, continuous on the real numbers. Regardless, it is an absolutely necessary statement and the teachers should understand this (even if they don't mention it to students at the time). The fact that this question was actually a discussion is, to be blunt, scary. $\endgroup$ – rnrstopstraffic May 12 '16 at 17:22
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The fallacy in the question is that (in $\mathbb R$) we do NOT

know $a\log x= \log x^a$

What actually holds is:$$x>0\implies a\log x=\log x^a$$

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Both equations are different and have different solutions. It matters whether or not the $x$ is squared initially. That's a subtle point. Any change to an equation makes it a different equation, technically.

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    $\begingroup$ That's not strictly true, and not a helpful thing to say. You need to make the distinction between uniquely reversible (one to one) changes and uniquely irreversible (many to one) changes. That's what's happening here - squaring is a many to one operation. $\endgroup$ – Grimxn May 11 '16 at 18:56
  • $\begingroup$ I'm not sure what you are saying. The fact is that different equations can have different solutions even if one equation can be derived from the other with simple operations. $\endgroup$ – jdods May 11 '16 at 18:59
  • $\begingroup$ "Can", not "must", which is the impression given by your answer. $\endgroup$ – Nij May 11 '16 at 20:18
  • $\begingroup$ There is nothing incorrect in my statement. In the situation under question there are 2 different equations with different solution sets. Nothing in my statement could be taken to mean that different equations always have different solution sets. Of course, your point really seems to be that this isn't a good answer, and certainly many of the other answers are better. $\endgroup$ – jdods May 11 '16 at 21:30

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