Trying to find the domain of this function.
I know that $\ln x$ available while $x>0$, I didn't knew from where to continue, any help appreciated.
i know that ln(x) available while x>0
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What values could $\ln(x)^2$ have? – J. W. Tanner Dec 16 '19 at 19:16
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@J.W.Tanner i assume negative and positive values but not 0 – Majd Dec 16 '19 at 19:17
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I mean if $x>0,$ what can you say about $\ln(x)^2$ (or anything squared)? – J. W. Tanner Dec 16 '19 at 19:18
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@J.W.Tanner positive – Majd Dec 16 '19 at 19:25
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Positive or zero, but if $\ln x=0$ (what’s $x$ then?) $\ln(\ln(x)^2)$ isn't defined – J. W. Tanner Dec 16 '19 at 19:29
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Define $$g(x)=\ln^2x$$ and $$f(x) = \ln x$$ Note that $$g(1) = \ln^2 1 = 0$$ Therefore, $$f(g(1)) = \ln 0$$ Which does not exist. You are also right saying that $\ln x$ is only defined for $x>0$. But if we define your function as $f(g(x))$ as we did above, we notice that $f(x)$ is only defined for $g(x) > 0$. Note that for all $x>0$, $g(x) \geq 0$, but equals $0$ at $g(1)$. We now have enough information to determine the domain of $f(g(x))$ as $$(0, 1) \cup (1, \infty) $$
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I can try to explain it better I guess. The domain of the inner function $ln^2(x)$ is $(0, \infty)$. The domain of the outer function $ln(x)$ is $(0, \infty)$. As a result, we would like to analyze the range of the inner function to see if it is ever less than or equal to $0$. As a matter of fact it does equal $0$ at $x=1$, so $1$ cannot be included in the domain of the entire function. The rest of the range is always greater than $0$ for the inner function, hence the final answer. Hope this helps. – Ty Jensen Dec 16 '19 at 19:41
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As you noted, $\ln x$ is defined (as a real function) only when $x>0$.
$\ln ((\ln x)^2$) is defined when $(\ln x)^2>0$.
For all $x>0$, $(\ln x)^2\ge0$, but $(\ln x)^2=0=\ln x$ when $x=1$.
Therefore, $\ln((\ln x)^2)$ is defined when $x>0$ and $x\ne1$.
J. W. Tanner
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