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Whenever I type some crazy big number on WolframAlpha, it outputs its power of 10 representation.

For example, if I input $$ \large e^{6^{3643.23976}}$$ it outputs $$ \large 10^{10^{2834.629359002706}}$$

What mathematical model allows WolframAlpha to calculate this? Is this an approximation or an exact representation of the number?

ihavenoidea
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    I'm guessing in this case this is doable because it uses the rule $a^b = c^{b \log_c a},$ in which case this should be exact. Just out of curiosity though, where are you getting this number from? Or are you just putting in big numbers for the fun of it. (which is also valid) – Stephen Donovan May 01 '21 at 17:26
  • @StephenDonovan It's pretty random, I was randomly browsing some highly upvoted questions here and stumbled on this answer which shows a similar number. I was curious about how big that number is. – ihavenoidea May 01 '21 at 17:30
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    It can’t be an exact representation. – Thomas Andrews May 01 '21 at 17:57
  • While Mathematica produces an Overflow[] result for E^6^3643.23976, the result for N[Log[10, Log[10, E^6^(364323976/10^5)]],16] is 2834.629359002706. – Somos May 01 '21 at 20:52
  • This is based on the simple formula for the double logarithm of a power $x^{y^z}$ , which is $z\cdot \lg(y)+\lg(lg(x))$ , where "$lg(x)$" denotes the logarithm with respect to base $10$. – Peter May 08 '21 at 15:18
  • This could be even calculated with a table calculator although the input itself is too large for every tool. – Peter May 08 '21 at 15:21

1 Answers1

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For this case, $x=e^{6^y}$ you get:

$$ \log_{10} x =6^y\log_{10} e$$

And:

$$\log_{10}\log_{10} x=y\log_{10} 6 +\log_{10}\log_{10}e\tag 1$$

Given any $y$ we can calculate approximately $z=\log_{10}\log_{10} x$ and then give a result of: $$x=10^{10^z}.$$

Note, it would be harder to do: $\log_{10}\log_{10}\log_{10} x$ in this case, because (1) is a sum. But if $y$ was large enough to make a third log necessary, you could treat the $+\log\log e$ part as zero.

Note, any error in your calculation of $z$ gives a potentially large error in your estimation of $x.$ The estimation of $x$ should be taken in the spirit of “orders of magnitude” rather than something more precise. If $z>1000,$ the number of digits of $z$ you’d need to know the number of digits for the integer part of $x,$ within $1,$ is very large. You’d need $z$ to the $1000$ place after the decimal, roughly.

Thomas Andrews
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