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This is the equation I am supposed to solve: $$\log_{3}(2x^{2} - x-1)\ -\ \log_{3}(\ x-1)\ = 2$$ The textbook gives the solutions, $x=1, x=4$, with the working out as shown below:

Working out of equation

While I understand what they are doing, I don't completely understand why they work it out as they do. Specifically, why isn't the term on the left side in the second row simplified? In my working out, $$\log_{3}\left(\frac{2x^{2} - x-1}{x-1}\right) \implies \log_{3}\left(\frac{(2x+1)(x-1)}{x-1}\right) \implies \log_{3}(2x+1)$$ This equals $2$, allowing it to be simplified further, \begin{align*} \: & \log_{3}(2x+1) = 2 \\ \implies \: & 2x+1 = 3^2 = 9 \\ \implies \: & 2x = 8 \\ \implies \: & x = 4 \\ \end{align*}

What's more, $x$ can't even equal $1$, because if it does, both logs simplify to $log(0)$, which is undefined. I am wondering if there is some other concept I am not aware of, or is the book making a mistake?

Patrick Yoder
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    When the book divides by $x-1$ it has to assume that $x\neq 1$. The only solution is $x=4$. – Kavi Rama Murthy May 21 '22 at 12:09
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    $x=1$ is not a solution . This fact is clearly correct. We can not divide to zero. Also $\log 0$ is undefined. – User May 21 '22 at 12:09
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    This point is under-emphasized in teaching algebra apparently. The implication $$ \log_{3}\left(\frac{(2x+1)(x-1)}{x-1}\right) \implies \log_{3}(2x+1) $$ is false, because the rational function has a "hole" in its domain at $ \ x \ = \ 1 \ \ . $ (The "cancelation" of factors "buries" this property of the original function.) People often don't write this, but it ought to be expressed explicitly that $$ \log_{3}\left(\frac{(2x+1)(x-1)}{x-1}\right) \implies \log_{3}(2x+1) \ , \ x \ \neq \ 1 \ \ . $$ Then the "extraneous" solution is already excluded. –  May 21 '22 at 23:14
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    I'll leave this as a comment, since other posters already discuss the problem extensively. I'll say that the solver for the textbook problem blithely multiplies both sides of the equation (potentially) by zero at one point: what should have been written, for the reason I've mentioned, $$ \frac{2x^2 \ - \ x \ - \ 1}{x \ - \ 1} \ \ = \ \ 9 \ \ \Rightarrow \ \ 2x^2 \ - \ x \ - \ 1 \ \ = \ \ 9x \ - \ 9 \ \ , \ \ x \ \neq \ 1 \ \ , $$ with the restriction continuing on all subsequent lines. (emacs below also points out that $ \ x \ = \ 1 \ $ isn't even in the domain of the original equation.) –  May 22 '22 at 07:35

4 Answers4

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You are right, $x=1$ is not a solution. Note that:

$$\log_ca-\log_cb=\log_c\left(\frac ab\right)$$

Holds iff. $0<c\neq 1$ and $a,b>0$.

This phenomenon is called the Extraneous root in mathematics.

FShrike
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User
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$x=1$ cannot be a solution because the equation is undefined for $x=1$. The equation in reals is only meaningful if $x-1 > 0$ and $2x^2-x-1>0$, which means $x>1$. In that case one may rewrite the equation as follows:

$$\begin{align} \log_3(2x^2-x-1) - \log_3(x-1) &= \log_3\big((2x+1)(x-1)\big) - \log_3(x-1) \\ &= \log_3(2x+1) + \log_3(x-1) - \log_3(x-1) \\ &= \log_3(2x+1) \\ &\stackrel!= 2 \end{align}$$ and thus $2x+1=9$ which has only one solution $x=4$.

As you see, the culprit is not a division by $x-1$ as clained in the comments, because it's still invalid if no division is present like above.

emacs drives me nuts
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The issue of whether the book is technically correct is (unfortunately) a subjective communication issue.

Technically, the statement $x=1$ or $x=4$ can be interpreted as implying that any value of $x$ that satisfies the equation must be an element of $\{1,4\}$.

Therefore, the set $\{1,4\}$ represents the two candidate values that must each be individually explored to determine which of them (if any) solve the problem.

So, under this interpretation, the book is merely saying that it is impossible for any value of $x$ to solve the problem if that value of $x$ is not an element in $\{1,4\}.$

Unfortunately, it is generally commonly understood that the naked statement $x=1$ or $x=4$ usually connotes that these are the answers, rather than that these are the candidate values.

So, the author was lazy, and just about any Math student (including me) would have looked at that and said : "on what planet"?

user2661923
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  • Very interesting. Knowing the context of the problem, as well as the fact that this book has made other blatant mistakes, it probably suits best to say this is simply another of them. Thanks again. – Patrick Yoder May 21 '22 at 13:14
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Technically, it is correct to write that \begin{align}&\log_3 (2x^2 - x-1)- \log_3(x-1) = 2\\\implies{}& x=1 \quad\text{OR}\quad x=4.\end{align} On the other hand, it would be incorrect to claim that the equation $$\log_3 (2x^2 - x-1)- \log_3(x-1) = 2$$ has solutions $1$ and $4,$ that is, that \begin{align}&\log_3 (2x^2 - x-1)- \log_3(x-1)\ = 2\\\iff{}& x=1 \quad\text{OR}\quad x=4.\end{align}

So, are you saying the book is technically correct? Maybe you could explain more in an answer?

  1. Yes, albeit incomplete. If the exercise is to solve that given equation and you've completely reproduced your book's answer, then it can be finished by filling in the missing final step: pruning out any extraneous solution, in this case $x=1$ as you've pointed out.

    Noting the attendant implicit conditions $$a\in(0,1)\cup(1,\infty)\quad\text{and}\quad x>0$$ of $\log_ax$ can help eliminate extraneous solutions.

  2. Alternatively (as you can see, the preceding pruning method is easier), its solution can be modified by carefully adapting the ⟹ to ⟺; here's a very detailed example: \begin{align}&\log_3 (2x^2 - x-1)- \log_3(x-1) = 2\\\iff{}&\log_3 \frac{2x^2 - x-1}{x-1} = 2 \quad\text{AND}\quad2x^2 - x-1>0\quad\text{AND}\quad x-1>0\\\iff{}&(\log_3(2x+1) = 2 \quad\text{AND}\quad x\ne1)\quad\text{AND}\quad2x^2 - x-1>0\quad\text{AND}\quad x-1>0\\ \iff{}& 2x+1= 9\quad\text{AND}\quad2x^2 - x-1>0\quad\text{AND}\quad x-1>0\\\iff{}& x=4.\end{align}

ryang
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