Fix $b>1, y>0$, and prove that there is a unique real $x$ such that $b^{x}=y$, by completing the following outline. (This $x$ is called the logarithm of $y$ to the base b.)
(a) For any positive integer $n, b^{n}-1 \geq n(b-1)$.
(b) Hence $b-1 \geq n\left(b^{1 / n}-1\right)$.
(c) If $t>1$ and $n>(b-1) /(t-1)$, then $b^{1 / n}<t .$
(d) If $w$ is such that $b^{w}<y$, then $b^{w+(1 / n)}<y$ for sufficiently large $n$; to see this, apply part $(c)$ with $t=y \cdot b^{-w}$
(e) If $b^{w}>y$, then $b^{w-(1 / m)}>y$ for sufficiently large $n$.
(f) Let $A$ be the set of all $w$ such that $b^{w}<y$, and show that $x=\sup A$ satisfies $b^{x}=y .$
(g) Prove that this $x$ is unique.