Logarithm overkill!

1. Jan 23, 2008

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?

2. Jan 23, 2008

mjsd

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

3. Jan 23, 2008

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

4. Jan 23, 2008

Shooting Star

Hint: the answer is an integer.

5. Feb 3, 2008

kuahji

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

6. Feb 3, 2008

tramtran111

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

7. Feb 3, 2008

danielle36

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

8. Feb 9, 2008

This is true

9. Feb 9, 2008

malawi_glenn

NEVER post the answer just like that!

10. Feb 9, 2008

Oerg

haha, I guess a perfect hint would be

$$lne^x=x$$