(a) For natural $n$, $b^n - 1 ≥ n(b-1)$.
(b) Hence $b - 1 ≥ n(b^{1/n}-1)$.
(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 $n$.
(e) If $b^w > y$ then $b^{w-1/n} > y$ for sufficiently large $n$.
(f) Let $A = \{w~|~w < y\}$ and show $b^{\text{sup }A} = y$.
(g) Show $x$ is unique.
Proof: (a) We show $b^n ≥ n(b-1)+1$. This is clear when $n=1$, so by induction$$b^n = bb^{n-1} ≥ b((n-1)(b-1)+1)) = (n-1)b(b-1)+b$$and we must show $(n-1)b(b-1)+b ≥ n(b-1)+1$. Collecting terms on the left and manipulating we must thus show $(n-1)(b-1)^2 ≥ 0$, which is clear as all these terms are nonnegative.
(b) This is clear from the previous, as $b^{1/n} > 1$ by noting $c^n ≤ 1$ for $c ≤ 1$ by induction.
(c) Assume $b^{1/n} ≥ t$. Then $b - 1 < n(t-1) ≤ n(b^{1/n}-1)$, a contradiction by (b).
(d) We have $1 < \dfrac{y}{b^{w}}$. Observe $b^{1/n} ≤ \dfrac{b-1}{n}+1$ by (b) so we may choose $n$ such that $\dfrac{b-1}{n}+1 < \dfrac{y}{b^{w}}$ by conditioning $\dfrac{b-1}{n} < \dfrac{y}{b^w}-1 > 0$ and now $n > \dfrac{b-1}{\dfrac{y}{b^w}-1}$ so that $b^{w+1/n} < y$.
(e) As before, since we showed $b^{1/n}$ tends toward 1 we can choose $n$ such that $b^{1/n} < \dfrac{b^w}{y}$ to fulfill.
(f) By (a) we have $b^w$ is divergent so $A$ has $x = \text{sup }A$. Suppose $b^x < y$; then by (d) we have some $b^{x+1/n} < y$ so $x$ is not an upper bound. Suppose $b^x > y$; then by (e) choose $n$ for $b^{x-1/n} > y$ and $x$ is not a minimal upper bound. Therefore $b^x = y$.
(g) If $x' \neq x$ then $b^x - b^{x'} = b^{x'}(b^{x-x'}-1) = 0$ is a contradiction.$~\square$
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