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I know that $ x \gt 0 $ because of logarithm precondition, and I can see that $ x \neq 1 $ because otherwise it would lead to $ 0^0$ which is problematic, but when I checked the graph of the function I have discovered that it started from $ 1 $ on $ x $ axis and $ (0,1) $ interval is not considered.

In sum, I thought that the domain should have been $ (0, \infty) - \{1\}$ but it seems to be $(1,\infty)$ and I can't figure out where I am wrong.

Emil Cohen
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2 Answers2

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Note: this question has only been addressed at the high school level. Therefore, the given answer does not address the question from a mathematician's point of view.

What I want to do in this answer is to find the largest interval, such that for all values ​​in this interval the function $f(x)=(\ln x)^{\ln^2 x}$ is well-defined/meaningful.

By definition we have $0<x\neq 1$.

$$\operatorname{dom}\left[(\ln x)^{\ln^2 x}\right]\longmapsto \ln x>0\vee \left\{\ln x<0\wedge \ln^2x\in\mathbb Z\right\}$$

and

$$\ln x>0\implies x>1$$

Then note that, if $\ln x<0$ then, $\ln ^2x \in\mathbb Z$.

Thus,

$$\ln x=-\sqrt n\implies x=e^{-\sqrt n},\, n\in\mathbb Z_{>0}.$$

With all this in mind, we can conclude that,

$$ \begin{aligned}&\operatorname{dom}\left[(\ln x)^{\ln^2 x}\right]\\ =&\left\{x\mid x>1\vee x=e^{-\sqrt n}, \,n\in\mathbb Z_{>0}\right\} \end{aligned} $$

If you assume $0^0=1$, then $x=1$ is included in the domain:

$$ \begin{aligned}&\operatorname{dom}\left[(\ln x)^{\ln^2 x}\right]\\ =&\left\{x\mid x\ge1\vee x=e^{-\sqrt n}, \,n\in\mathbb Z_{>0}\right\}\end{aligned} $$

nonstudent
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  • Is the first $\land$ in your first line supposed to be $\lor$? – user170231 Aug 30 '22 at 17:22
  • @user170231 I confused a bit... I must use "or" instead of "and"? – nonstudent Aug 30 '22 at 17:30
  • I believe so. $$\ln x>0\wedge \left{\ln x<0\wedge \ln^2x\in\mathbb Z\right}$$ sounds contradictory to me - how can both $\ln(x)>0$ and $\ln(x)<0$ and $\ln^2(x)\in\Bbb Z$? I think you meant to say $$\ln x>0\vee \left{\ln x<0\wedge \ln^2x\in\mathbb Z\right}$$ – user170231 Aug 30 '22 at 20:46
  • @user170231 Thank you for reporting this mistake. Now, I will edit the answer. – nonstudent Aug 30 '22 at 20:53
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First of all, we can see that $f(e^{-1}) = {(\ln e^{-1})}^{\ln^{2}e^{-1}} = -1$ so we can see that it is indeed defined for some values of $x < 1$ but $f(e^{-1/2})$ is undefined for example.

If you think about it, this question is very similar to asking for the domain of $x^x$ so this might help you. Furthermore, as @David said on his comment, if $a < 0$ then $a^b$ is "problematic". Knowing this, to avoid any problems we would want $\ln x > 0 \implies x > 1$ so we could just conveniently define $\text{dom} f(x) = (1,+\infty)$. And in this case:

$$f(x) = {(\ln x)}^{\ln^{2}x} = (e^{\ln(\ln x)})^{\ln^2 x} = e^{\ln(\ln x) \cdot \ln^2 x}$$

Now let's try to find a larger domain that includes some $x < 1$, to avoid raising numbers to an irrational exponent that is another problem on its own, let's consider all rational $\ln x < 0$, then for some $p,q \in \mathbb{N}$, $0 <x = e^{p/q} < 1$ and thus $$f(x) = f(e^{p/q}) = (p/q)^{(p/q)^2} = \sqrt[q^2]{(p/q)^{p^2}}$$ remember that $p/q < 0$ so for this expression to be properly defined we want $p$ to be even or $q$ to be odd.

So we can as well define the domain of $f(x)$ as $(1,+\infty) \cup \{\ e^{p/q} \in (0,1) \mid p = 2k \text{ or } q = 2k + 1 \text{ for some } k\in\mathbb{Z} \}$.

So its domain really depends on what you define it to be.

Torrente
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