# What is the domain of f(x)=ln(ln(ln(x)))?

1. Sep 13, 2007

### frasifrasi

Can anyone explain how to get the domain of f(x)=ln(ln(ln(x)))?

I am stuck at e^lnlnx > e^o...

Thank you.

2. Sep 13, 2007

### learningphysics

simplify the left side... and the right side is just 1.

3. Sep 13, 2007

### arildno

We must have ln(ln(x))>0
Hence, we must have
ln(x)>1

whereby you should readily find the domain of x!

4. Sep 13, 2007

### frasifrasi

Can someone explain the steps, I REALLY don't get this and it has been bothering me for a week.

5. Sep 13, 2007

### bomba923

As arildno said, determine what values of the inner functions belong to the domains of the outer functions: $$\ln \ln \ln x \in \mathbb{R} \Rightarrow \ln \ln x > 0 \Rightarrow \ln x > 1 \Rightarrow x > e$$

6. Sep 13, 2007

### Dick

Ok, as this has been going on for two threads now. For ln(ln(ln(x))) to be defined, the argument of the leftmost ln must be > 0. So ln(ln(x))>0. Exponentiate both sides. exp(ln(ln(x)))=ln(x), exp(0)=1. Hence ln(x)>1 since exp(x) is monotone. Exponentiate again. exp(ln(x))=x. exp(1)=e. So x>e. Since exp(x) is monotone still. This is exactly the same thing everyone else has been saying. What don't you 'get'?

7. Sep 14, 2007

### learningphysics

e^(lnlnx) = lnx (because e^(lnr) = r for any r)

lnx>e^0

lnx>1 (since e^0 = 1)

Therefore

e^(lnx) > e^1 (take e to the power of both sides)

we know that e^(lnx) = x (same reason as before)

so substitute that into the previous inequality...

x > e^1

x > e

8. Sep 14, 2007

### frasifrasi

Thank you physics!!!!!!!!!!1