How do you solve ${ln x}^{2} = 4$ $ln x^2=4$ ?

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Answer 1 · 2015-07-24T19:22:53

$x \in {- e^{2}, e^{2}}$ $x in {-e^2, e^2}$

${ln x}^{2} = 4$ $lnx^2=4$

$\Rightarrow x^{2} = e^{4}$ $=>x^2=e^4$

$\Rightarrow x^{2} - e^{4} = 0$ $=>x^2-e^4=0$

Factorize,

$\Rightarrow (x - e^{2}) (x + e^{2}) = 0$ $=>(x-e^2)(x+e^2)=0$

There are two solution,

$\Rightarrow x - e^{2} = 0 \Rightarrow x = e^{2}$ $=>x-e^2=0 =>x=e^2$

And,

$\Rightarrow x + e^{2} = 0 \Rightarrow x = - e^{2}$ $=>x+e^2=0=>x=-e^2$

How do you solve lnx2=4ln x^2=4?