Is ln|x|/x equal to ln|x^(-x)|?

[find_the_fun]

Is [math]\frac{\ln{|x|}}{x}=\ln{|x^{-x}|}[/math] because of the rule [math] y \ln{x}=\ln{x^y}[/math]?

 

[Dethrone]

Yep! And in general, you can bring the exponent inside the absolute value.

 

[find_the_fun]

Yep! And in general, you can bring the exponent inside the absolute value.

Oh cool, usually my answer key makes these sorts of simplifications but it didn't for this one.

 

[I like Serena]

Is [math]\frac{\ln{|x|}}{x}=\ln{|x^{-x}|}[/math] because of the rule [math] y \ln{x}=\ln{x^y}[/math]?

Applying that rule gives us:
$$\frac{\ln{|x|}}{x}=\ln{\left(|x|^{1/x}\right)}$$
And I'm afraid we can't generally bring that power inside the absolute signs.