# Natural Logarithm Laws

#### 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

Homework Helper
How does
(ln(x))^(1/x)=ln(x^(1/x))?
It doesn't, in general. It does if x = 1.

Homework Helper

#### 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(ab)=(b)ln(a)

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

Last edited:

#### Towk667

General formula: ln(ab)=(b)ln(a)

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.

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