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

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The equation $\frac{\ln{|x|}}{x}=\ln{|x^{-x}|}$ is examined in relation to the rule $y \ln{x}=\ln{x^y}$. While the rule allows for simplification, it does not permit bringing the exponent inside the absolute value in all cases. The transformation leads to $\frac{\ln{|x|}}{x}=\ln{\left(|x|^{1/x}\right)}$, but this does not equate to $\ln{|x^{-x}|}$. The discussion highlights the importance of understanding the limitations of logarithmic properties when dealing with absolute values. Overall, the simplification does not hold universally.
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Is [math]\frac{\ln{|x|}}{x}=\ln{|x^{-x}|}[/math] because of the rule [math] y \ln{x}=\ln{x^y}[/math]?
 
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Yep! And in general, you can bring the exponent inside the absolute value.
 
Rido12 said:
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.
 
find_the_fun said:
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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