Help with Comparing k1 < k2 in Homework Problem

[Gear2d]

Homework Statement

k is a constant where k₁ < k₂

Is the relation between the two <. > or = to:

A = n^k₁ + k₂ⁿ

B = n^k₂ + k₁ⁿ

The Attempt at a Solution

I did this problem said that A > B, and I got it wrong. I am having a hard time telling if n^k_i or k_iⁿ is greater.

 

[jhicks]

Do you mean *n* is a constant? I don't see a k. Also, you have given us insufficient information to solve the problem because depending on k1 and k2 relative to n the answer will change. For example, n=10, k1=1, k2=2 and n=2, k1=10, k2=11.

 

[Dick]

If n=1 then A>B. If n=2, k1=1 and k2=2 then A>B. If n=2, k1=3 and k2=4, then A<B. Do you know something about n you aren't telling us? What makes you think there is a definite relation between the two?

 

[Gear2d]

Sorry about that, n can be any number, does not have to be fixed like the k's. So, for example n^k₁ < n^k₂.

 

[Dick]

Sorry about that, n can be any number, does not have to be fixed like the k's. So, for example n^k₁ < n^k₂.

That doesn't help. You've already been given examples where A<B and B>A.

 

[Ad2d]

I think that k's are fixed value for a give problem, and that as n-> infinite that one of those will be >, <, or =. I think that is what gear2d is asking.

 

[Dick]

I think that k's are fixed value for a give problem, and that as n-> infinite that one of those will be >, <, or =. I think that is what gear2d is asking.

You might be right. If that's the real question, I wish Gear2d would clarify.

 

[Gear2d]

Thanks Ad2d you said what I wanted to say. So how would one approach this problem?

 

[Dick]

Figure out which term is dominant. Take n^k1 and k2^n. The logs are k1*log(n) and n*log(k2). Which is larger as n gets large? I.e. what is lim n->infinity (k1*log(n))/(n*log(k2))? Is it zero or infinity? Think l'Hopital.

 

[Gear2d]

Thank you, that makes sense. Don't know why I did not see that. Thanks again.