Solve Logarithm Overkill: Find ln(ln[e^{e^{5}}])

[danielle36]

[SOLVED] Logarithm overkill!

Hello again!
I have been working on this log, and the longer I work on it, the more confused I get! Here's the problem:
Find the exact value for:

ln(ln[e^{e^{5}}])

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Here's what I've tried so far:

e(ln[e^{e^5}}])

e^{x} = ln(e^{e^5}})
e^{x} = e^{e^5}}
e^{5} = (2.72)^{5}
e^{x} = e^{149}
x = 149

...I have no idea if I'm doing this right, but I'm not feeling like I am...Help?

 

[mjsd]

are you trying to do two logs then two e's or one e then a log then two e's
very confusing!

 

[ahmadmz]

You have to use one of the properties of logs. When the bases of the exponent and log are same, they cancel.

 

[Shooting Star]

Hint: the answer is an integer.

 

[kuahji]

ln e ^x =x
Can you take it from there?

 

[tramtran111]

i think the answer is 5 since ln and e cancaled out!

 

[danielle36]

hey thanks everyone! i was able to figure it out from there you guys are always a big help :)

 

[Hydrargyrum]

i think the answer is 5 since ln and e cancaled out!

This is true

 

[malawi_glenn]

i think the answer is 5 since ln and e cancaled out!

NEVER post the answer just like that!

 

[Oerg]

NEVER post the answer just like that!

haha, I guess a perfect hint would be

lne^x=x