# Baby Rudin - Existence of logarithm

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• Aug 30th 2010, 12:56 PM
Math Major
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 \$?
• Aug 31st 2010, 01:03 AM
Opalg
Quote:

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.
• Aug 31st 2010, 03:59 AM
Math Major
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!