What is the domain and range of $y = ln (x^{2})$ $y=ln(x^2)$ ?

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What is the domain and range of $y = ln (x^{2})$ $y=ln(x^2)$ ?

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Answer 1 · 2017-04-09T14:02:19

Domain for $y = ln (x^{2})$ $y=ln(x^2)$ is $x \in R$ $x in R$ but $x \neq 0$ $x!=0$ , in other words $(- \infty, 0) \cup (0, \infty)$ $(-oo,0)uu(0,oo)$ and range is $(- \infty, \infty)$ $(-oo,oo)$ .

Explanation:

We cannot have logarithm of a number less than or equal to zero. As $x^{2}$ $x^2$ is always positive, only value not permissible is $0$ $0$ .

Hence domain for $y = ln (x^{2})$ $y=ln(x^2)$ is $x \in R$ $x in R$ but $x \neq 0$ $x!=0$ , in other words $(- \infty, 0) \cup (0, \infty)$ $(-oo,0)uu(0,oo)$

but as $x \to 0$ $x->0$ , $ln (x^{2}) \to - \infty$ $ln(x^2)->-oo$ , $y$ $y$ can take any value from $- \infty$ $-oo$ ao $\infty$ $oo$ i.e. range is $(- \infty, \infty)$ $(-oo,oo)$ .

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What is the domain and range of $y = ln (x^{2})$ $y=ln(x^2)$ ?

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