# Comparing $\large 3^{3^{3^3}}$, googol, googolplex

How to show that $\large 3^{3^{3^3}}$ (Third Ackermann number) is larger than a googol ($\large 10^{100}$) but smaller than googoplex ($\large 10^{10^{100}}$).

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Please do not deface the question. It orphans the answers that people have given. –  robjohn Feb 11 '13 at 18:33

$$3^{3^{3^3}} > 3^{300} > 10^{100}$$ 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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@user60465, dont try to do that edit - it ruins my post. –  user58512 Feb 11 '13 at 17:35

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 $27 < 100.$

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