Hi $\displaystyle ln(x^{2}) $ is defined for all $\displaystyle x \neq 0 $ But, using basic logarithm law: $\displaystyle ln(x^{2}) = 2 \cdot ln(x) $ ???
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Originally Posted by Twig Hi $\displaystyle ln(x^{2}) $ is defined for all $\displaystyle x \neq 0 $ But, using basic logarithm law: $\displaystyle ln(x^{2}) = 2 \cdot ln(x) $ ??? ln(x) is still undefined for x=0 so I'm not sure what you're trying to ask
Originally Posted by Twig $\displaystyle ln(x^{2}) $ is defined for all $\displaystyle x \neq 0 $ But, using basic logarithm law: $\displaystyle ln(x^{2}) = 2 \cdot ln(x) $ It works like this: $\displaystyle \sqrt {x^2 } = \left| x \right|$. Thus, $\displaystyle \ln \left( {x^2 } \right) = 2\ln \left( {\left| x \right|} \right)$.
Let $\displaystyle f(x) = ln(x^{2}) $ Then, $\displaystyle D_{f}=\mathbb{R} $ , except $\displaystyle x = 0 $ But, by writing $\displaystyle 2 \cdot ln(x) $ I dont have the same domain anymore. Now I need $\displaystyle x > 0 $
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