# Is the function $y=\ln x^2$ the same as $y=2\ln |x|$?

Suppose I have a function $$y=\ln x^2$$ Then is this function the same as $$y=2\ln |x|?$$

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Short answer: yes. – Javier Jul 15 '12 at 16:49
Yes it is.   – Did Jul 15 '12 at 16:49
...if $x\in\mathbb R$, of course. – J. M. Jul 15 '12 at 16:52
$$\log x^2 = \frac 1 2 2 \log x^2 = 2\log\sqrt{x^2}=2\log |x|$$ – Pedro Tamaroff Jul 15 '12 at 22:17
Writing $\ln x^2$ is ambiguous, it may mean $(\ln x)^2$. – GEdgar Jul 15 '12 at 23:50

A function has both a domain and a rule which tells you what to do with elements in the domain. If we are assuming $x \in \mathbb{R}$, then both functions listed have domain all reals other than 0. And, since $2 \ln |x| = \ln |x|^2 = \ln (x^2)$, these two functions have the same rule for all $x$ in their identical domains. So, they are the same function.

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As many have shown, if $x$ is real the functions are the same. However, if $x$ is a complex number the result is not true.

In fact if $z$ is complex and nonzero $$\log(z^2) = 2\log|z| + i \mathrm{Arg}(z^2) + 2\pi n i$$ where $\mathrm{Arg}(z^2)$ is the principal value of the complex argument of $z^2$ and where $n=0,\pm1,\pm2,\ldots$.

If $z$ is real and we choose the principal branch of the logarithm then $$\log(z^2) = 2\log|z|$$ since in this case $\mathrm{Arg}(z^2) = \mathrm{Arg}\,1 = 0$ and $n=0$.

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Over $\Bbb R,$ the question whether $\ln |x|^2 \overset{?}{=} \ln x^2$ is essentially $|x|^2 \overset{?}{=} x^2.$ The proof is trivial, but oh well:

Proposition: For all $x \in \Bbb R$ we have $|x|^2 = x^2.$

Proof: If $x \ge 0,$ then $|x| = x,$ and $|x|^2 = x^2.$ If $x < 0,$ then $|x| = -x,$ and $|x|^2 = (-x)^2 = x^2.$

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Yes, indeed $$\ln(x^2) = 2\ln \lvert x \lvert .$$ First though we have (as proved in J.D.'s answer) that $x^2 = \lvert x \lvert^2$. And so \begin{align} \ln(x^2) &= \ln(\lvert x\lvert^2) \\ &= 2\ln\lvert x \lvert. \end{align}

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