Natural Logarithm Laws

**Towk667** · Nov16-09, 02:27 AM

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** · Nov16-09, 02:34 AM

Originally Posted by Towk667

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

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

**HallsofIvy** · Nov16-09, 06:38 AM

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

**HallsofIvy** · Nov16-09, 06:46 AM

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

**Towk667** · Nov16-09, 12:14 PM

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 $LaTeX Code: \\infty0$ 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** · Nov16-09, 04:30 PM

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** · Nov16-09, 04:48 PM

Originally Posted by 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.

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

**g_edgar** · Nov17-09, 09:14 AM

Originally Posted by Towk667

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

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