on internet we usually see: $$2\log(x) = \log(x^2) $$ (example)but how is this true? one is defined for $x>0$ and the other one for $x\neq0$
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8One ought to write $\log (x^2)=2\log |x|$. – lulu Dec 11 '21 at 16:14
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4That identity holds for positive real x, in the same way as $\sqrt{x^2} = x$ holds only for $x \ge 0$, or $1/(1/x) = x$ holds only for $x \ne 0$. – Martin R Dec 11 '21 at 16:15
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4Fun fact: $1\ne\frac{x}{x}$ – Soumil Gupta Dec 11 '21 at 16:17
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You can have two functions defined on different domains, which coincide on the intersection of their domains. – Martin R Dec 11 '21 at 16:18
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4Does this answer your question? Is Wolfram Alpha calculating this incorrectly? – Martin R Dec 11 '21 at 16:26
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In the reals , this identity is valid since it only makes sense anyway in the case $x>0$. We cannot compare something that is not defined with anything else (not even with something undefined as well). – Peter Dec 11 '21 at 16:32
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The assertion $\log(x^2)=2\log(x)$ and similar expressions usually means that we have that equality when both the LHS and RHS are defined. It's like the equality$$\frac1{1/x}=x.\tag1\label1$$The LHS is undefined when $x=0$, whereas the RHS is defined for every number. And asserting that $\eqref1$ holds means, in this case, that it holds when $x\ne0$.
Joe
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José Carlos Santos
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thank you, sorry but this was in one of the prof solutions and he passed from $log((t-1)^2)$ to $log((1-t)^2)$ – Dec 11 '21 at 18:24