Given n > 8, let a = √n and b = √(n+1). Which is greater a^{b} or b^{a}?

**Solution**

Straightforward

Answer: a^{b} is greater.

a^{b} = e^{b ln a} and b^{a} = e^{a ln b}. So we have to decide which of b ln a and a ln b is greater, or, equivalently, which of (ln a)/a and (ln b)/b is greater. The latter is clearly more promising. So set f(x) = (ln x)/x. Then f '(x) = 1/x^{2} - (ln x)/x^{2} which is negative for x > e. Obviously b > a, so provided a > e, (ln a)/a > (ln b)/b and hence b ln a > a ln b and a^{b} > b^{a}. But e^{2} < 9, so the result is certainly true for n ≥ 9.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999