$e^{e^{10^{10^{2.8}}}}$ changing $e$ with $10$

Question

From Numberphile $$e^{e^{10^{10^{2.8}}}}$$ changing $e$ with $10$, is there a way to change only the top most number while keeping all other numbers 10? i.e what is x in :

$$e^{e^{10^{10^{2.8}}}} = 10^{10^{10^{10^{x}}}} $$

Answer 1

In general, not necessarily. The left-hand side needs to be big enough. The smallest we can get is if we let $x \to -\infty$, which gives us $$ 10^{10^{10^{10^{x}}}} \to 10^{10^{10^{10^{-\infty}}}} \to 10^{10^{10^{0}}}=10^{10^{1}} = 10^{10} $$ so basically, we cannot reach down to $10^{10}$, but any number above that is large enough that we can find an $x$ that takes us there. Clearly, $e^{e^{10^{10^{2.8}}}}$ is large enough.

Following Dave's comment above, we can find an (approximate) $x$ the following way (letting $\ln$ be the natural logarithm and $\log$ the base-ten logarithm): $$ e^{e^{10^{10^{2.8}}}} = 10^{10^{10^{10^{x}}}}\\ e^{10^{10^{2.8}}} = 10^{10^{10^{x}}}\ln 10\\ 10^{10^{2.8}} = 10^{10^{x}}\ln(10) + \ln(\ln 10) \approx 10^{10^x}\ln(10)\\ 10^{10^{2.8}} \approx 10^{10^x + \log(\ln 10)}\\ 10^{2.8} \approx 10^x + \log(\ln 10)\\ x \approx 2.7997506116 $$ so we actually don't have to change the exponent much. The approximation of throwing away $\ln(\ln 10) \approx 0.83$ happens at a stage where both sides have $630$ digits, so it doesn't change much. The $\log(\ln 10)$, however, is then the only thing left that separates the two cases, so we have to keep that, and at that stage, the two sides have value at around $630$ instead, so it has a much greater impact on the result anyways.

Answer 2

First, we have

$$ e^{\textstyle 10^{10^{2.8}}} = 10^{\textstyle \log_{10}(e) \cdot 10^{10^{2.8}}} = 10^{\textstyle 10^{\left( 10^{2.8} + \log_{10}\log_{10}(e) \right)}} = 10^{\textstyle 10^{10^{A}}} $$ where $$A = \log_{10}(10^{2.8} + \log_{10}\log_{10}(e)) \approx \log_{10}(630.96 - 0.36) \approx 2.7998 $$

Now, setting $B=10^{10^A}$, do the same once more: $$ e^{\textstyle e^{\textstyle 10^{10^{2.8}}}} = e^{\textstyle 10^B} = 10^{\textstyle \log_{10}(e)\cdot 10^B} = 10^{\textstyle 10^{B+\log_{10}\log_{10}(e)}} $$ But $\log_{10}\log_{10}(e)$ is still only $-0.36$, whereas $B$ has hundreds of digits before the decimal point, so this equals $$ 10^{\textstyle 10^{\textstyle 10^{10^C}}} $$ for a $C$ that is so close to $A\approx2.7998$ that it isn't even funny.

Moral: The values of the bottommost bases in a power tower matter very little in the great scheme of things.