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Mathematics 21 Online
OpenStudy (anonymous):

How do you find the domain of f(x)=lnlnlnx

OpenStudy (turingtest):

I know you know it sat...

OpenStudy (turingtest):

x>e^e right?

OpenStudy (anonymous):

domain of \[ln(x)\] is \[x>0\] so domain of \[\ln(\ln(x))\] is \[\ln(x)>0\] \[x>1\] lather rinse repeat

OpenStudy (anonymous):

i think you might be off by one @turing

OpenStudy (turingtest):

d'oh!

OpenStudy (anonymous):

For ln(x), x>0. We can agree on this, right? For no n can e^n=<0. However, ln(x) can come out to be ln(x)<0, unless x>e (otherwise for e^n n<0). So the domain of ln(ln(x)) must have x>e.

OpenStudy (anonymous):

did i mess up? let me check

OpenStudy (anonymous):

no i don't think so

OpenStudy (turingtest):

no you guys both agree I was wrong

OpenStudy (anonymous):

\[\ln(\ln(\ln(x)))\] must have \[\ln(\ln(x))>0\] must have \[\ln(x)>1\] must have \[x>e\]

OpenStudy (anonymous):

Woo!

OpenStudy (anonymous):

Thanks guys!!!

OpenStudy (anonymous):

Omg i actually understand it! Thank you, saved me a good hour of wasting time going to my school's tutoring center.

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