# Thread: Baby Rudin - Existence of logarithm

1. ## Baby Rudin - Existence of logarithm

I'm working out of Baby Rudin, chapter 1 problem 7. The problem is "Fix b >1, y > 0, and prove that there is a unique real x such that $\displaystyle b^x = y$ by completing the following outline..."

I'm at part f) "Let A be the set of all w such that $\displaystyle b^w < y$, and show that $\displaystyle x = sup(A)$ satisfies $\displaystyle b^x = y$.

What I'm wondering about - if m is a positive integer, since b > 1, is it a leap to assume that there's an integer k such that $\displaystyle b^k > m$?

2. Originally Posted by Math Major
I'm working out of Baby Rudin, chapter 1 problem 7. The problem is "Fix b >1, y > 0, and prove that there is a unique real x such that $\displaystyle b^x = y$ by completing the following outline..."

I'm at part f) "Let A be the set of all w such that $\displaystyle b^w < y$, and show that $\displaystyle x = sup(A)$ satisfies $\displaystyle b^x = y$.

What I'm wondering about - if m is a positive integer, since b > 1, is it a leap to assume that there's an integer k such that $\displaystyle b^k > m$?
You can easily enough justify that "leap" by writing b = 1 + h, where h > 0. Then $\displaystyle b^k = (1+h)^k > 1 + kh$ (since the right-hand side is just the first two terms of the binomial expansion), and that can clearly be made arbitrarily large.

3. Oh wow, thank you. Never would've thought of the binomial theorem. I kept trying in my head to figure out a way to write k without referencing a logarithm >.<. Thanks!