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Prove without using a calculator $$(\ln 6)^{(\ln 5)^{(\ln 4)^{(\ln 3)^{(\ln 2)}}}}<\pi$$ I want to know if there is an easy way to prove this inequality without using a calculator.

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  • $\begingroup$ To what base are you taking your logarithms? $\endgroup$ – Mark Bennet Oct 23 '15 at 11:25
  • $\begingroup$ the base is $e$ or I mean $\log x=\ln x$ $\endgroup$ – E.H.E Oct 23 '15 at 11:26
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    $\begingroup$ May I ask where this problem is from ... interestingly (with a calculator) the difference is quite small. $\endgroup$ – martini Oct 23 '15 at 11:30
  • $\begingroup$ it looks like a chance rather than having a reason. It it not correct for when you extend it to number 7. $\endgroup$ – Arashium Oct 23 '15 at 11:56
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    $\begingroup$ whats the motivation for doing such a question? $\endgroup$ – user210387 Oct 23 '15 at 11:58
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I don't know how to prove it, yet. However, I can help motivate a reason to solve it.

As to your question,

"I want to know if there is an easy way to prove this inequality without using a calculator?"

The answer is no. There is no easy/quick proof of this that will make it's way into an answer. However, I have made some observations.

We have a recursion relation of the following,

$$L(x+1)=\ln(x+1)^{L(x)}$$

$$L^{(0)}(\ln(x))=\ln(x) \quad L^{(1)}(\ln(x))=\ln(x+1)^{\ln(x)}$$

$$\ln(23)=3.13549...$$ $$L^{(1)}(\ln(6))={\ln(7)}^{\ln(6)}=3.29623...$$ $$L^{(2)}(\ln(4))={\ln(6)}^{{\ln(5)}^{\ln(4)}}=\color{green}{3.08961...}$$ $$L^{(3)}(\ln(3))={\ln(6)}^{{\ln(5)}^{{\ln(4)}^{\ln(3)}}}=\color{blue}{3.16664...}$$ $$L^{(4)}(\ln(2))={\ln(6)}^{{\ln(5)}^{{\ln(4)}^{{\ln(3)}^{\ln(2)}}}}=\color{red}{3.14157...}$$ $$L^{(5)}(\ln(1))={\ln(6)}^{{\ln(5)}^{{\ln(4)}^{{\ln(3)}^{\ln(2)^{\ln(1)}}}}}=\color{blue}{3.16664...}$$ $$L^{(6)}(\ln(0))={\ln(6)}^{{\ln(5)}^{{\ln(4)}^{{\ln(3)}^{\ln(2)^{\ln(1)^{\ln(0)}}}}}}=\color{green}{3.08961...}$$

I find that interesting and motivating.

Now, some of you might be intrigued by colorful use of $\ln(0)$ above. I actually found that using a limit, we observe the behavior of $L^{(6)}(\ln(x))$ as $x$ approaches $0$.

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    $\begingroup$ $\ln(0)$? Something I'm missing? ;-) $\endgroup$ – egreg Oct 23 '15 at 15:17

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