Given the following:


If I were to use the power rule, I would do:






Substituting back into the original equation, it is correct. However, there is a missed answer:


Substituting that into the original equation, it is correct as well, but while using the power rule, it has been overlooked.

Obviously, this is a rather simple question, so I could have just done $3^2=(x^2-3)^2$ and solved it like that, in which case the answer of $0$ would appear. However, for more complicated questions, how would one go about doing it? And are there exceptions to this exception?

  • $\begingroup$ Just as a little side question, how does one preserve answers in the following:$$\log(x^n)$$where we have $x\in\mathbb{C}$? $\endgroup$ Jan 28, 2016 at 0:43

5 Answers 5


The lost step was the firs one: $\log_3({x^2-3})^2=2\log_3(|x^2-3|)$ (always that $|x^2-3|\neq0$. In your particular problem, always that $x\neq \pm\sqrt{3}$).

Now, if $|x^2-3|\neq 0$, then $2=\log_3({x^2-3})^2=2\log_3(|x^2-3|)$ iff $1=\log_3(|x^2-3|)$ iff $3=|x^2-3|$ iff $x=0$ or $x=\pm\sqrt{6}$

  • $\begingroup$ Can u please explain where the mod comes from? Thanks! $\endgroup$
    – Max Payne
    Jan 27, 2016 at 12:18
  • $\begingroup$ @Tim: What "mod"? $\endgroup$ Jan 27, 2016 at 17:38
  • 1
    $\begingroup$ The mod is precisely cause $|y|^2=y^2$. $\endgroup$
    – sinbadh
    Jan 27, 2016 at 17:41
  • 2
    $\begingroup$ interesting question! sneaky appearance of domain restrictions. I suppose in retrospect the flag to look for here is that the log expression is a quartic polynomial, so we should be on the lookout for 4 solutions (0 being a "repeated root") $\endgroup$ Jan 27, 2016 at 20:22

So, the problem arises because of this fact: $\text{log}(a^b)=b\text{log}(a)$ only if $a>0$. So, when you brought the 2 down using the power rule, you were assuming that $x^2-3>0$, when now that you see one solution is $x=0$ so it need not be the case that $x^2-3>0$. That is where the zero solution got thrown out. So in this case, changing it to exponential form retains all the solutions.

With more complicated problems, you just have to keep in mind whether or not the rule you want to use will throw out solutions. I hope this helps!


In this case, if you want to preserve the domain ($\mathbb{R}\setminus\{\pm\sqrt3\}$), you have $$ \log_3|x^2-3|=1, $$ which gives the additional solution.


There is NO 'exception to the power rule' here, it's your error in expression handling. The error is similar to saying, if $(x+1)^2 > 9$ then $x+1>3$.

The square in the inner function guarantees the logarithm argument is non-negative, and the equation's domain is all reals with two points excluded: $$\{x:(x^2−3)^2 > 0\} = \{x:(x^2−3)\ne 0\} = \mathbb R \setminus \{\pm\sqrt 3\}$$ 'Removing' the square cuts-off the domain portion in which $(x^2-3)<0$, leaving $$\{x:(x^2-3)>0\} = \mathbb R \setminus [-\sqrt 3,\sqrt 3]$$ hence you miss solution(s) inside the interval $(-\sqrt 3,\sqrt 3)$.


I think the exception of the power rule exists because (let $a$ equal the following) $$ a = x^2-3 $$ is not strictly >=0, as per x=0, you'll see that it equals negative 3, so taking the logarithm of something like that is undefined.

$a^2$ is strictly non-negative, hence the logarithm can be solved as you mentioned properly with all the solutions, so once you use the power rule, it breaks the 'strictly non negative'-ness of logs.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .