What is the Correct Form of the Natural Logarithm Law for (ln(x))^(1/x)?

[Towk667]

How does
(ln(x))^(1/x)=ln(x^(1/x))?

A friend told me this was a true statement but could'nt prove it. If that isn't true, then how would you find the lim x->0 of (ln(x))^(1/x) using L'Hospital's Rule?

 

[CRGreathouse]

How does
(ln(x))^(1/x)=ln(x^(1/x))?

It doesn't, in general. It does if x = 1.

 

[HallsofIvy]

For example, if x= 2, ln(2)= 0.69315, approximately so (ln(2))^{1/2}= 0.83255. But 2^{1/2}= 1.41421 so ln(2^{1/2})= 0.34657. Not at all the same.

 

[HallsofIvy]

For example, if x= 2, then ln(2)= 0.69315, approximately, and (ln(2))^{1/2}= 0.83255.<br /> <br /> But 2^{1/2}= 1.41421 and so ln(2^{1/2})= 0.34657. Not at all the same.&lt;br /&gt; &lt;br /&gt; As for the entire problem of finding the limit, as x goes to 0, of (ln(x))^{1/x}, I see a serious difficulty: as soon as x&amp;lt; 1, ln(x)&amp;lt; 0 and fractional powers of negative numbers are not defined.

 

[Towk667]

That's what I thought, but my friend insisted that it was true. I've been rattling my brain for about 2 days on that one, so I decided to ask here. So can you help me with limit I mentioned in my first post? I typed it wrong in the first post its the limit as x approaches infinity not zero. I can see from graphing it that it's going to come out to one, but I don't know how to use L'Hopistal's Rule to solve for it. If I try to evaluate it without changing anything I get something like \infty⁰ which would be one if it isn't indeterminant, I don't remember if it is or isn't. Anyways, I'm supposed to use L'Hosp. Rule and I don't know how to write the limit as a fraction to use L'Hopistal's Rule though.

 

[mathman]

General formula: ln(a^b)=(b)ln(a)
For your formula: ln(x^1/x)=(1/x)ln(x)

As for the L'Hopital rule question, you don't need it, since the expression goes to (-∞)^∞, which is ∞, with an ambiguous sign.

 

[Towk667]

General formula: ln(a^b)=(b)ln(a)
For your formula: ln(x^1/x)=(1/x)ln(x)

As for the L'Hopital rule question, you don't need it, since the expression goes to (-∞)^∞, which is ∞, with an ambiguous sign.

The original equation is [ln(x)]^(1/x) not ln(x^(1/x)).

 

[g_edgar]

The original equation is [ln(x)]^(1/x) not ln(x^(1/x)).

...and the original equation was incorrect, so mathman gave something correct.