How to show that $\large 3^{3^{3^3}}$ is larger than a googol ($\large 10^{100}$) but smaller than googoplex ($\large 10^{10^{100}}$).
Thanks much in advance!!!
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2Please do not deface the question. It orphans the answers that people have given. – robjohn Feb 11 '13 at 18:33
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$3^{3^{3^{3}}}$ is not the third Ackermann number. The latter is $3\uparrow^{3}3 \ = \ 3\uparrow^{2}(3\uparrow^{2}3))$ (an exponential tower of 7625597484987 threes). I've edited the question accordingly. – r.e.s. Apr 02 '14 at 21:31
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@robjohn, plus, that edit would make no sense. I don't no why the author did it. – JMCF125 Apr 04 '14 at 15:36
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$$3^{3^{3^3}} > 3^{300} > 10^{100}$$ since $$3^{3^3} > 3^7 = 3 \cdot (3^3)^2 > 3 \cdot 10^2 = 300$$ since $$3^3 > 7$$
$$3^{3^{3^3}} < 10^{3^{3^3}} < 10^{10^{100}}$$ since $$3^{3^3} < 3^{100} < 10^{100}$$ since $$3^3 < 100$$
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I know others have beat me by nearly a day, but here's something I came up with just now that seems more straightforward. Each of the inequalities makes frequent use of monotonicity (increasing), either in the base or in the exponent, of an exponentiated expression. (After writing this up, I noticed that my estimates in carrying out the googolplex part are the same as what user58512 has.)
$$10^{100} \; < \; {\left( 3^3 \right)}^{100} \; < \; {\left( 3^3 \right)}^{\left( 3^5\right)} \; = \; 3^{\left( 3 \cdot 3^5\right)} \; = \; 3^{3^{6}} \; < \; 3^{3^{3^3}} $$
In the strict inequalities above, I first made use of $10 < 3^3,$ then $100 < 3^5,$ and lastly $6 < 3^3.$
$$3^{3^{3^3}} \; < \, 10^{3^{3^3}} \; < \; 10^{10^{3^3}} \; = \; 10^{10^{27}} \; < \; 10^{10^{100}}$$
In the strict inequalities above, I first made use of $3 < 10,$ then $3 < 10,$ and lastly $27 < 100.$
Dave L. Renfro
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\begin{align} &\color{red}{100 \log_3 10} < 100\times3 < 729 = 3^6 < \color{red}{3^{3^3}} = 3^{27} < 10^{100} < \color{red}{10^{100} \log_3 10}\\ \Rightarrow&10^{100} < 3^{3^{3^3}} < 10^{10^{100}}. \end{align}
user1551
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