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Could we say it has one when substituting the $x$ value into the original equation? Obviously, to solve the trascendental equation one has to operate the two terms with logarithms, reducing the equation to $\log((x-10)/(x-3))=3$ and finally having $x$ approximately equal to $2.633$. Now how can we explain a student that the value obtained, if substituted into the original equation, yields both a LHS and a RHS not in reals?

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    $\begingroup$ Recall extraneous solutions to radical equations in school algebra. The situation here is essentially the same, in the sense that certain algebraic manipulations can take non-solutions to solutions. Examples: (1) $-2$ is not a solution to $x=2,$ but if you square both sides, then we get $x^2 = 4,$ which has $-2$ as a solution. (2) $0$ is not a solution to $x=2,$ but if you multiply both sides by $x,$ then we get $x^2 = 2x,$ which has $0$ as a solution. $\endgroup$ – Dave L Renfro Feb 18 at 15:33
  • $\begingroup$ Absolutely so. Jordan Calmes and Ananth Jayadev say in Brilliant.com that "A solution to an algebra problem is valid if both sides of the equation are still equal when the problem has been worked out with the chosen solution substituted for the variable(s)". $\endgroup$ – Tavasanis Feb 18 at 15:37
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    $\begingroup$ One of the pitfalls that can come up with logarithm manipulations and identities (indeed, even square root manipulations), is overlooking restrictions for the manipulations and identities to be valid. For example, in using $\log ab = \log a + \log b,$ the typical restriction is that $a > 0$ and $b > 0.$ The problem is that when you're applying this identity where at least one of the logarithm inputs involves an unknown $x,$ you often don't know in advance whether the value(s) you ultimately arrive at for $x$ are such that the inputs to the logarithms are positive. So you have to verify it. $\endgroup$ – Dave L Renfro Feb 18 at 15:38
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    $\begingroup$ Therefore, the moral we as teachers should always have is to mind restrictions whenever this kind of trascendental equations appear... $\endgroup$ – Tavasanis Feb 18 at 15:43
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These kinds of problems typically show up (in the US, at least) in precalculus classes, hence appeals to the derivative or branches of the complex logarithm, while correct, miss the mark (in my opinion). My presumption is that by the time a student knows any calculus or complex analysis (1) they should know better (or, at least, they should know that they should know better; if you remind them that the logarithm of a negative number is not real, they should immediately recognize that there is an issue) and (2) these kinds of problems are a bit artificial—while issues with the domains of logarithms pop up, they rarely do so in the manner presented in the question (what is much more common, I think, is for sign errors to crop up when the argument of the logarithm is smaller than $1$).

When working through such a problem with students in precalculus classes, I tend to remain vigilant throughout the process regarding the restrictions on the possible values of $x$. For example, in the given problem, I would explain as follows:

Find all real solutions (if any) to the equation $$ \log(x-10) = 3 + \log(x-3), $$ where $\log : (0,\infty) \to \mathbb{R}$ is the real natural logarithm

By definition, the logarithm is a function on the positive real numbers. Thus this equation only "makes sense" if $x-10 > 0$ and if $x-3 > 0$. That is, we are trying to find a real number $x$ which has the following three properties:

  1. $\log(x-10) = 3 + \log(x-3)$,
  2. $x - 10 > 0$, and
  3. $x - 3 > 0$.

Conditions 2 and 3 simplify to $x > 10$: since $x - 10 < 0$, it follows from additive cancelation (that is, if follows from "adding $10$ to both sides") that $x > 10$. In other words, $x \in (10, \infty)$. Similarly, it must also be the case that $x > 3$, and so $x \in (3, \infty)$. As both conditions must be satisfied, $x$ must be in both intervals, which means that $x$ is in the intersection: $$ x \in (10,\infty) \cap (3, \infty) = (10,\infty). $$ Hence $x > 10$.[1]

Therefore \begin{align} &\log(x-10) = 3 + \log(x-3) &\text{and} &&& x > 10 \\ &\qquad\iff \log\left(\frac{x-10}{x-3}\right) = 3 &\text{and} &&& x > 10 \\ &\qquad\iff \frac{x-10}{x-3} = \mathrm{e}^3 &\text{and} &&& x > 10 \\ &\qquad\iff x-10 = (x-3)\mathrm{e}^3 &\text{and} &&& x > 10 \\ &\qquad\iff x - \mathrm{e}^3 x =(1-\mathrm{e}^3) x = 10 - 3\mathrm{e}^3 &\text{and} &&& x > 10 \\ &\qquad\iff x = \frac{10 - 3\mathrm{e}^3}{1 - \mathrm{e}^3} \approx 2.633 &\text{and} &&& x > 10. \end{align} It is not possible for this final condition to be satisfied, therefore the original equation has no real solution.


[1] In the precalculus class that I generally teach, some very elementary set theory and the notation for intervals is established early on. In a course where this is not the practice, the discussion of Conditions 2 and 3 can be handled in whatever manner is appropriate for the class, e.g. shaded pictures of number lines, other algebraic techniques, etc. The point is to determine restrictions on the possible values of $x$ as early as possible in the process.

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No, you can see that $\log(x-10) < 3 + \log(x-3)$ by checking the derivatives: $$0 < \frac{1}{x-10} < \frac{1}{x-3}$$ and the value at any point where they are both defined. i.e. $\log(x-10)$ starts below $3+\log(x-3)$ and grows slower.

Of course, if you either expand the domain by rewriting the equation as $$ \log | x-10 | = 3+\log|x-3| $$ or by allowing $\log$ to take negative real inputs at the expense of throwing off extra $\pi\sqrt{-1}$ terms, then $$x = \frac{3e^3 - 10}{e^3 - 1}$$ becomes a solution. In the absolute value case, we actually pick up yet another solution, as you can see in the graph below.

Graph of log|x-10| and 3+log|x-3|

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  • $\begingroup$ Yours is a decisive proof. Thank you so much. $\endgroup$ – Tavasanis Feb 18 at 15:40
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    $\begingroup$ by checking the derivatives --- Simpler (don't have to take derivatives and then use the racetrack principle) and more elementary (don't need to use calculus methods) is to use the fact that $\log x$ is a (strictly) increasing function. Therefore, since $(x-10) < (x-3),$ we have $\log (x-10) < \log (x-3).$ Also, $\log (x-3) < 3 + \log (x-3).$ Therefore, it follows that $\log (x-10) < 3 + \log (x-3).$ $\endgroup$ – Dave L Renfro Feb 18 at 20:19
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The way I explain it is that the log rule that you're using - the claim that

$$\log_ax - \log_ay = \log_a\frac{x}{y}$$

is only correct if $\log_ax$ and $\log_ay$ both exist. This is pretty clear if you prove the rule from the relevant exponent law - you have to start by writing $x$ and $y$ as powers of $a$, which can't be done if they're nonpositive. So, essentially, in the process of solving we made a mistake by using a rule that wasn't correct. Now, one way to fix it would be to be more careful: every time you use the rule, add a caveat of "unless $x \leq 0$" or what have you, and deal with that as a separate case. But that's needlessly complicated, so the way we usually approach it is to just pretend that the rule always works, and let the verification process tell us when it doesn't.

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  • $\begingroup$ I like writing unless X<=0 more, since it prompts to consider the domain of existence of the functions involved and is generally a good habit. However, probably, not all students will find it intuitive $\endgroup$ – May Feb 28 at 23:04
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Title question: Uh...yes. Approximately 2.633.

"Now how can we explain a student that the value obtained, if substituted into the original equation, yields both a LHS and a RHS not in reals?"

By just showing them? It's not a requirement that intermediate parts of some other equation are real. Just that x is.

But consider, that you could also think of the problem as having started with the log of the fraction. Thus breaking it out into the separate logs would be the manipulation creating the strange result. Maybe there are many ways to generate such oddities, starting from something non-weird.

FWIW: You can also have real roots of cubics that depend on complex numbers in their intermediate solution.

P.s. I hope your students are well-drilled on the basics. I see many questioners/answers at this site who are fascinated by the "oolies" (harder problems, definitional distinctions) while simultaneously teaching a lot of kids who really need better drill in the basics.

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  • $\begingroup$ Good answer, thanks. I only emphasized the presentation of the original equation, which, as is, suggests the restriction $x > 10$ to be real on both sides, while the solution is much lower. $\endgroup$ – Tavasanis Feb 18 at 15:03
  • $\begingroup$ Good point. (extra characters) $\endgroup$ – guest Feb 18 at 15:07
  • $\begingroup$ P.s. If you are using log to mean base 10 (not to get into that debate, but showing my applied bias), the answer is a rational number. $\endgroup$ – guest Feb 18 at 15:12
  • $\begingroup$ If you enter the equation in Symbolab, for example, they show the whole process until the solution is found, but they say the equation has no solution in $\mathbb{R}$ because the last verification step results false. $\endgroup$ – Tavasanis Feb 18 at 15:14
  • $\begingroup$ With natural log base, the result is trascendental because it involves $e$, and in other bases may be rational, but I think it's out of the point as regards verification. $\endgroup$ – Tavasanis Feb 18 at 15:16

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