Is $\frac{\ln{|x|}}{x}=\ln{|x^{-x}|}$? Explained by y $\ln{x}=\ln{x^y}$ Rule

[find_the_fun]

Is \(\displaystyle \frac{\ln{|x|}}{x}=\ln{|x^{-x}|}\) because of the rule \(\displaystyle y \ln{x}=\ln{x^y}\)?

 

[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 \(\displaystyle \frac{\ln{|x|}}{x}=\ln{|x^{-x}|}\) because of the rule \(\displaystyle y \ln{x}=\ln{x^y}\)?

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.

 

[CellCoree]

No, the two expressions are not equivalent. The first expression, $\frac{\ln{|x|}}{x}$, is a logarithmic function with a variable base of $e$ and a variable exponent of $x$. The second expression, $\ln{|x^{-x}|}$, is a logarithmic function with a base of $e$ and an exponent of $-x$. The two expressions have different bases and exponents, therefore they are not equal.

The rule $y\ln{x}=\ln{x^y}$ only applies when both the base and exponent are constants. In this case, $x$ is a variable, so the rule cannot be used to equate the two expressions. It is important to carefully consider the properties of logarithmic functions and their rules before attempting to equate expressions involving them.