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PhiNotPi's avatar

How to tell the greater power? (math question)

Asked by PhiNotPi (12681points) December 20th, 2011

You have two numbers: the first is 45362^15272, and the second number is 26453^37589. How can I figure out which one is the largest without actually calculating the full value of each number? I am not interested in which is the largest out of my example, I want to find the strategy to figure out which is the largest.

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6 Answers

BonusQuestion's avatar

Compare the logs?

Taking log from the first one you get 15272 log(45362) and from the second one you get 37589 log(26453) which can be compared easily.

PhiNotPi's avatar

@BonusQuestion What would that easy comparison consist of? In the example, it is obvious (one is greater than one, one is less than one), but that may not always be the case.

roundsquare's avatar

@PhiNotPi How is one greater than 1 and the other less than 1? The log of each is between 4 and 5.

But I think @BonusQuestion is right. Once you have a log(b) and c log(d) you look at orders of magnitude of b and d (i.e. which two integers the log’s are between) and see if that helps.

In general, you can also just use orders of magnitude. So in your example:

(Note: I’m using ~=~ to mean “approximately equal to”).

45362^15272 = (4.5362*10^4)^15272 = 4.5362^15272 * 10^31088 ~=~ 10^10000 * 10^31088 = 10^41088

And so something similar for the second one. Then compare the exponent.

I think if you got good at doing this it would be fast.

PhiNotPi's avatar

@roundsquare @BonusQuestion I misinterpreted the logs. I thought that they were to be considered as one number was the base of the log, while the other was what we were taking the logarithm of. If that was the case, one would be less than one and one greater, and the comparison would be easy. I can see what is meant by them now, and they would be between four and five.

ratboy's avatar

45362 ~ 26453^1.052961.
45362^15272 ~ 26453^(1.052691*15272) ~ 26453^16077.
16077 < 37589.

Given A^n and B^m with A < B, calculate log_A(B) = log(B)/log(A).
Compare n with log_A(B)*m.

finkelitis's avatar

Here’s a simple viewpoint: the power of one is more than twice the power of the other. So you can rewrite:

26453^37589 = ((26453)^2)^[half of 37389]

Clearly 26453 squared is greater than 45362, and half of 37389 is greater than 15272. So you end up with a larger number raised to a larger power. No logs, no calculators.

Similarly, you can use some estimation to make the problem even simpler to handle:

26453^37589 > 20000^36000 = (20000^2)^18000 = (400000000)^18000 > 45362^15272.

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