Solving Logarithm Overkill: Find Exact Value for ln(ln[e^{e^{5}}])

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To find the exact value of ln(ln[e^{e^{5}}]), one can simplify the expression step by step. First, recognize that ln[e^{e^{5}}] simplifies to e^5, as ln and e are inverse functions. Then, applying the natural logarithm again, ln(e^5) equals 5. Therefore, the exact value of ln(ln[e^{e^{5}}]) is 5. This method effectively utilizes logarithmic properties to arrive at the solution.
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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?
 
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use the property that Ln(a^b) = b Ln(a), Ln(e) = 1, Ln(e^a) = a
 
z=ln(lne^(e^5)). Use the definition of a log now.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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