How to compare two huge numbers with nested exponentials?

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Discussion Overview

The discussion revolves around comparing two extremely large numbers represented as nested exponentials, specifically in the form a^b^c^d and w^x^y^z. Participants explore the feasibility of using logarithmic transformations to compare these numbers without calculating their actual values, addressing the challenges posed by the nested structure of the exponentials.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using logarithms to compare the two nested exponentials, noting that for simpler cases, logarithmic transformations can simplify comparisons.
  • Another participant questions whether the proposed transformation log(a^b^c) = b^c log a can be further simplified when b^c is still large, indicating uncertainty about the effectiveness of this approach.
  • There is a discussion about whether applying logarithms consecutively can help eliminate nested exponentials, with one participant expressing doubt about the utility of taking the logarithm of sums.
  • Participants engage in correcting each other's mathematical expressions, indicating a collaborative effort to refine their understanding of the logarithmic properties involved.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the effectiveness of using logarithms to compare the nested exponentials. There are multiple competing views regarding the validity of the proposed transformations and the potential for simplification.

Contextual Notes

The discussion highlights limitations in the proposed methods, particularly regarding the handling of large values and the application of logarithmic properties to sums and products. There is an unresolved nature to the mathematical steps involved in simplifying the nested exponentials.

xopek
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Let's say we have two numbers represented as a "tower" of exponentials, a^b^c^d and w^x^y^z (powers calculated right to left) and we want to compare them, not necessarily calculating their values. Their values are so huge, they can't be represented on a computer or calculator. Is it possible to use logarithms to compare them? I know that it is possible for a simple case, say, a^b and x^y. We can apply log to both sides and then compare b log a and y log x. But what about nested powers? We can represent log(a^b^c) as b^c log a. But what if b^c is still huge. Can we continue and end up with something like c* log(b) * log(a)? A quick test shows that this is probably not going to work. Any ideas? Thanks
 
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xopek said:
...We can represent log(a^b^c) as b^c log a. But what if b^c is still huge. Can we continue and end up with something like c* log(b) * log(a)? A quick test shows that this is probably not going to work. Any ideas? Thanks
log(bc⋅log(a)) = c⋅log(b)+log(log(a)) , not what you have.
 
SammyS said:
log(bc⋅log(a)) = c⋅log(b)+log(log(a)) , not what you have.
Ah, right, thank you. So do you think the general approach is otherwise correct?

Let's say in c⋅log(b)+log(log(a)) c is not just c but to the power of d. I am not sure, would I need to apply log to the sum? In other words, is there a general approach to get rid of all the nested exponentials by applying log consecutively? I am not sure what to do here:
log( cd⋅log(b)+log(log(a)))
 
Last edited:
xopek said:
Ah, right, thank you. So do you think the general approach is otherwise correct?

Let's say in c⋅log(b)+log(log(a)) c is not just c but to the power of d. I am not sure, would I need to apply log to the sum? In other words, is there a general approach to get rid of all the nested exponentials by applying log consecutively? I am not sure what to do here:
log( cd⋅log(b)+log(log(a)))
Once you have that sum, taking the log doesn't help.
 

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