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This is concerning a question in stack exchange : Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$. I was actually wondering why the correct result is not obtained when applying $\log$ on both sides,like how the second answer does.Why is this so? Is it possible to get this answer using logarithms?

Asaf Karagila
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Hema
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2 Answers2

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Yes, you are right! We can use logarithm here: $$(x^2+4x-60)\ln(x^2-5x+5)=0,$$ which gives $$x^2-5x+5=1$$ or $$x^2+4x-60=0.$$ By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^{x^2+4x-60}$$ then $x^2-5x+5>0$ by definition.

Michael Rozenberg
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  • Can you please explain '$x^2-5x+5 > 0$ by definition'? – Toby Mak Oct 27 '18 at 07:31
  • @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that? – user Oct 27 '18 at 07:34
  • @Toby Mak If we want to work with expressions like $\left(f(x)\right)^{g(x)}$ then we need $f(x)>0.$ – Michael Rozenberg Oct 27 '18 at 07:34
  • @gimusi Try to calculate $\lim\limits_{x\rightarrow2}(x^2-5x+5)^{x^2+4x-60}.$ I think now you'll understand me. – Michael Rozenberg Oct 27 '18 at 07:37
  • @MichaelRozenberg I had in mind a solution like that one for x=2 we obtain $(-1)^{-52}=\frac1{(-1)^{52}}=1$. – user Oct 27 '18 at 07:42
  • @gimusi Yes $(-1)^{-52}=1$, but my limit does not exist. If we want to work with this expression we need $x^2-5x+5>0$. It's a definition only, but it's a very comfortable definition. – Michael Rozenberg Oct 27 '18 at 07:46
  • I agree with you and when we deal with calculus issues it is often a necessary limitation but here for an algebraic problem plugging x=2 we obtain the equality therefore I think that assuming that as a solution shouldn’t be considered a mistake. – user Oct 27 '18 at 07:50
  • @gimusi I think this situation must be impossible in math. What is it: sometimes we think that the domain of $(x^2-5x+5)^{x^2+4x-60}$ it's ${x|x^2-5x+5>0}$ and sometimes we don't think so? – Michael Rozenberg Oct 27 '18 at 07:57
  • @MichaelRozenberg I mean that in the algebraic perspective we can see at $$(x^2-5x+5)^{x^2+4x-60}=1$$ as an identity between expression without any needing of the concept of function and we can find others solution for $x\in \mathbb{N}$. I don't claim that it is a correct or official option but of course it is a reasonable option. – user Oct 27 '18 at 08:00
  • @gimusi I don't agree with you! We have a function here. We can not ignore this. Say me please. What is a value of the expression $\frac{x}{\sqrt{x-1}}$ for $x=0$? – Michael Rozenberg Oct 27 '18 at 08:02
  • @MichaelRozenberg That's a completely different case Michaeland I like do not agree with you sometimes! :) – user Oct 27 '18 at 08:05
  • @MichaelRozenberg To explain better my point, the question is not find $a$ such that $\lim_{x\to a} f(x)^{g(x)}=1$ but find x such that $f(x)^{g(x)}=1$ and for the latter $x=2$ works. (Note that also the tag is precalculus) – user Oct 27 '18 at 08:10
  • @gimusi It does not change something. Is algebra-precalculus not math? If we want to get something in math we need to use a language, which helps to make it. – Michael Rozenberg Oct 27 '18 at 08:14
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    @MichaelRozenberg Indeed it is math and $(-1)^{-52}=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :) – user Oct 27 '18 at 08:17
  • @gimusi Yes, of course! It's a definition only, but I think it's a very comfortable definition. I wrote in my post that it's my opinion only. – Michael Rozenberg Oct 27 '18 at 08:21
  • @MichaelRozenberg We agree on that, it is of course a useful and covenient convention/definition in calculus context but in other context it could limit the number of solutions we can find and usually we don't like that! That's nice, I'll take a look to your editing! – user Oct 27 '18 at 08:24
  • when you take logarithm, is not the rule: $(x^2+4x-60)\ln|x^2-5x+5|=0$, which may be the cause of problem, no? – farruhota Oct 27 '18 at 09:00
  • @farruhota I think it wold be another problem. – Michael Rozenberg Oct 27 '18 at 09:03
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The solution you have mentioned use indeed logarithm, that is for $A>0$

$$A^B=1 \iff \log A^B=\log 1 \iff B\cdot \log A=0 \iff B=0 \quad \lor \quad \log A=0$$

Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case

  • $A=-1$
  • $B=2k$ with $k\in \mathbb{Z}$

and include also the case $x=2$ among the solutions since $$(-1)^{-52}=\frac1{(-1)^{52}}=1$$

To summarize in order to address your question:

  • $A^B=1$ can be solved by logarithm only for the case $A>0$
  • $A^B=1$ has also one "special" solution for $A\le 0$ which cannot be find by logarithm but directly by the given conditions
user
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