calculate value of $2^{3^{4^{5}}}/e^{10240}$

Question

I am trying to calculate the value of $b=\dfrac{2^{3^{4^5}}}{e^{10240}}$. Is there any method to solve this efficiently?

Answer 1

As you can see WolframAlpha can give you an approximative power of 10 representation: $$10^{10^{10^{2.688465006395752}}}.$$

It is hard to get a better solution, because this number is too large. To show it, we will calculate the number of digits in base-$10$ representation. If $n$ is a number in base-$b$, with $d$ digits, then we know, that the formula for $d$ is the following: $$d=\lfloor\log_{b}n\rfloor+1.$$ Now let $$n:=\dfrac{2^{3^{4^5}}}{e^{10240}}.$$ Then the number of digits of $n$ in base-$10$ is $$d = \lfloor \log_{10} \dfrac{2^{3^{4^5}}}{e^{10240}}\rfloor+1 \\ d = \lfloor \log_{10} 2^{3^{4^5}} - \log_{10} e^{10240}\rfloor+1 \\ d = \lfloor 3^{4^5} \cdot \log_{10} 2 - 10240 \cdot \log_{10} e\rfloor+1.$$

We get the following for $d$:

$1124021466074751860097567522104789648012545442387518261576295420518 \\5174447660807915950553426138321488548657928846792570107753324167422 \\0100211779337007726069891241143955562498313809792175538301726950271 \\2513614070749429457547548532111853109636637797579524872247171419406 \\3487219461564568491620652987627661309480232956516340085351404053765 \\2037205369420431855146383193275981445894731731211119067826441631620 \\7609542700946643046958255703325110043123352486373327969799306832787 \\29227794366058969345$

So $n$ has $1.2402146607475186 \cdot 10^{489}$ digits.

You can easily check the result with Maple:

floor(3^(4^5)*log10(2)-10240*log10(exp(1)))+1;

This $n$ number is larger then a googleplex, which number also doesn't have a base-$10$ representation, because that is also too large. On the other hand this is less then Skewes' number and much much less then Graham's number. Read more at Large numbers wikipedia article.

The problem is that $2^{3^{4^5}}$ is too large. It is much larger then the denominator. The numerator and the whole fraction have just $4$ order of magnitude difference in base-$10$, the difference of the lengths is $4447$. The above give approximation has $1.124020417 \cdot 10^{489}$ digits, so the real value and the approximation has $1.049 \cdot 10^{483}$ difference in the numbers of digits.

Answer 2

You can use Wolfram Alpha for the question in the title, or here for the one in the body. Either one is enormous.