# Find the domain of the following function $f(x)={(\ln x)}^{\ln^{2}x}$.

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.

• If $a<0$ then $a^b$ can also be "problematic"... Aug 28, 2022 at 18:53
• the function $f(x) = a^x$ is defined if $a > 0$ then $\ln x$ must be $>0$.. Aug 30, 2022 at 12:16
• You cannot take real power of a négative number Aug 30, 2022 at 12:48
• @Lelouch: Well, you can, it just requires the use of complex numbers, and doesn't necessarily have a unique value.
– Dan
Aug 30, 2022 at 16:01

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 $$p,q \in \mathbb{N}$$, $$0 and thus $$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.

• Consider $f(x)=x$, then write $f(x)=e^{\ln x}$. Does this imply $\text{dom} f(x)=(0,+\infty)$? Aug 30, 2022 at 10:27
• @nonstudent You are right, just fixed the answer. Aug 30, 2022 at 12:26

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.

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

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}

• Is the first $\land$ in your first line supposed to be $\lor$? Aug 30, 2022 at 17:22
• @user170231 I confused a bit... I must use "or" instead of "and"? Aug 30, 2022 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 $$\ln x>0\vee \left\{\ln x<0\wedge \ln^2x\in\mathbb Z\right\}$$ Aug 30, 2022 at 20:46
• @user170231 Thank you for reporting this mistake. Now, I will edit the answer. Aug 30, 2022 at 20:53