# Baby Rudin Excercie 1.7 (Existence of Logarithms)

I'm currently working through Baby Rudin and need help on exercise 1.7(f).

The question:

Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x = \text{sup}(A)$ satisfies $b^x = y$.

Things given and proven in parts (a)-(e).

1. $b > 1$, $y > 0$, $x \in \mathbb{R}$
2. $\forall n \in \mathbb{Z}^+$, $b^n - 1 \geq n(b-1)$
3. $b-1 \geq n(b^{1/n}-1)$
4. $b^{1/n} < t$
5. $b^w < y \implies b^{w+(1/n)} < y$, for sufficiently large $n$, and using the substitution $t = yb^{-w}$.
6. $b^w > y \implies b^{w-(1/n)} > y$

I'm having trouble proving this. My approach so far is to try to prove that both $b^x < y$ and $b^x > y$ are false, so that $b^x = y$.

Proof (so far):

By definition, $\forall w \in A$, $x \geq w$. Assume $b^x < y$. $b^{x+1/n} < y \implies b^xb^{1/n} < y$ (This is where my proof falls apart, I'd like to use fact #4 to substitute in $t$, but it might invalidate my inequality. The same problem occurs if I assume $b^x > y$. Any ideas?)

well you certainly have $$b^x \le y$$ this suggests an approach based on
suppose $b^x \lt y$
can you use the facts you cite to show that $$\exists z \gt x\\ b^z \lt y$$ a contradiction?