B How to compare two huge numbers with nested exponentials?

[xopek]

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

 

[SammyS]

...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(b^c⋅log(a)) = c⋅log(b)+log(log(a)) , not what you have.

 

[xopek]

log(b^c⋅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( c^d⋅log(b)+log(log(a)))

 

[SammyS]

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( c^d⋅log(b)+log(log(a)))

Once you have that sum, taking the log doesn't help.