Quick Natural log/exp question

[frasifrasi]

A part of an example on my book goes from the following:

e^(lnlnx) > e^0

to

(e^lnx)^lnx > e^0

and to

ln x > e^0

this last step is what threw me off--shouldn't it be x^lnx instead of ln x on the right side since they are inverse functions. Is anything being done that I have missed?

Thank you.

 

[Dick]

This first line simplifies to ln(x)>0, the second line to x^ln(x)>0. They aren't the same.

 

[frasifrasi]

Ok, so let me ask the full question and see if it helps:

- Differentiate f(x) = lnlnlnx and find its domain.

So, after applying the chain rule twice, the answer is: (1/lnlnx)*1/lnx*1/x

Since the domain is the tricky part, can anyone explain how to get it?

According to the answer key, it is supposed to be x > e following the steps on my original post.

Thank you.

 

[frasifrasi]

Anyway, dick can you explain why the first line simplifies to lnx > 0? I don't get that either...

 

[Dick]

Anyway, dick can you explain why the first line simplifies to lnx > 0? I don't get that either...

e^(ln(a))=a. Now put a=ln(x).

 

[Dick]

Ok, so let me ask the full question and see if it helps:

- Differentiate f(x) = lnlnlnx and find its domain.

So, after applying the chain rule twice, the answer is: (1/lnlnx)*1/lnx*1/x

Since the domain is the tricky part, can anyone explain how to get it?

According to the answer key, it is supposed to be x > e following the steps on my original post.

Thank you.

What's the domain of ln(x)?

 

[frasifrasi]

no, the domain of (1/lnlnx)*1/lnx*1/x, which is the answer for the question.

 

[Dick]

no, the domain of (1/lnlnx)*1/lnx*1/x, which is the answer for the question.

The question asks for the domain of f(x)=lnlnlnx, not the domain of the derivative. They are different. I was just trying to get you to tell me that the domain of ln(x) is x>0. This tells you that ln(lnx)>0 (concentrate on the leftmost ln in the definition of f). Now where to go? Where is ln(ln(x))>0?

 

[frasifrasi]

I see what you mean, the language is vague there, but the professor most definitely wants the domain of the derivative...

I see what you mean that e^lnlnx is not the same as (e^ln)lnx since the first is a composite function and the latter is simply a product (power of power)--I guess the answer key was wrong about this, but can you explain what the derivative is anyways...

 

[Dick]

You already know the derivative, you told me. The hardest thing to satisfy in the domain of the derivative is that ln(ln(x)) be defined and non-zero. For what values of x is that true? You seem to understand everything else well enough I think you can tell me, and I won't have to break my blood oath not to reveal answers. :)

 

[frasifrasi]

For that anwer, i originally thought it couldn't be 1 or 0 because if x is 0, the domain is 0 and also that it couldn't be 1 because ln 1= 0. But, the prof. answer's sheet says x > e, so I am utterly perplexed. If anyone can explain the domain for that derivative, I would highly appreciate.

 

[Dick]

The domain of the derivative is x>1. And I expect you to explain to me why. Pronto. If the profs answer sheet says x>e then I think the organism in question wants the domain of f(x)=ln(ln(ln(x))). You aren't helping me enough to understand what you don't understand here.