How to compare two huge numbers with nested exponentials?

In summary: You need to apply it to each individual term, not the entire sum. This approach may not work for all cases, but it is a starting point for comparing nested exponential values represented as logarithms.
  • #1
xopek
24
0
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
 
Mathematics news on Phys.org
  • #2
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.
 
  • #3
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:
  • #4
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.
 

Related to How to compare two huge numbers with nested exponentials?

1. How do I compare two huge numbers with nested exponentials?

To compare two huge numbers with nested exponentials, you can use the logarithmic function. Take the logarithm of both numbers and compare the resulting values. The larger value will correspond to the larger number.

2. Can I use a calculator to compare these numbers?

Yes, most scientific calculators have a logarithmic function that you can use to compare two numbers with nested exponentials. Simply input the two numbers, take the logarithm, and compare the resulting values.

3. What if one number has a larger base than the other?

If one number has a larger base than the other, you can use the change of base formula to convert both numbers to the same base. Then, take the logarithm and compare the resulting values.

4. Are there any other methods for comparing these numbers?

Another method for comparing two numbers with nested exponentials is to rewrite both numbers in exponential form with the same base. Then, compare the exponents. The larger exponent will correspond to the larger number.

5. Can I compare these numbers without using logarithms?

Yes, you can also compare two numbers with nested exponentials by simplifying both expressions. This can be done by expanding the exponents and then comparing the resulting terms. However, this method may be more time-consuming and may not work for extremely large numbers.

Similar threads

Replies
1
Views
1K
Replies
4
Views
2K
Replies
19
Views
3K
Replies
10
Views
2K
Replies
10
Views
1K
  • General Math
Replies
4
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
Replies
1
Views
789
Replies
1
Views
1K
  • Computing and Technology
Replies
4
Views
799
Back
Top