Prove the uniqueness of $x$ such that $b^x=y$ This is a follow up a different part of the question I asked here from Rudin exercise 7, chapter 1: Prove the uniqueness of $x$ such that $b^x = y$ (part a)

Fix $b > 1$, $y > 0$, and prove that there is a unique real $x$ such that $b^x = y$, by completing the following outline.
(a) For any positive integer $n$, $ b^n-1 \geq n(b-1) $

This follows from the binomial theorem, $$b^n-1 = (b-1)(b^{n-1} + b^{n-2} + \dots + b + 1) \geq (b-1)n$$ and
$$(b^{n-1} + b^{n-2} + \dots + b + 1) \geq n$$ because b > 1.

(b) Hence $b - 1 \geq n(b^{1/n} - 1)$

Since $b > 1$ and $n > 0$, we know that $(b^{1/n}-1)$ will be 0 or negative. Hence $b - 1 \geq n(b^{1/n} - 1)$. However now that I am re-writing this on stackexchange, this is not true because $(b^{1/n}-1)$ is not always have less than one and always negative so my reasoning isn't good enough here :(

(c) If $t > 1$ and $ n > (b-1)/(t-1)$, then $b^{\frac{1}{n}} < t$

Here, Since $n$ is a positive integer, it would make sense that $b^{1/n} < t$ if $b = t$, but I am not sure how to be more general to account for, if for example, $t = 10, b = 144,$ and $n = 1/2$, then $ 144^{1/2} < 10$ which would be false.
Same reasoning goes for thinking of $b < t^n$, well if $b=10, t=2,$ and $n=3$, then $10 < 2^3$ is false too.
I am not sure how I can think of these problems differently, but I feel that I definitely need to consider more things to get proper answers.
 A: For part (b), define
$$ c = b^{1/n}. $$
Note that $b > 1$ implies that $c > 1$ (you should be able to reason about why this is true without too much difficulty).  Then, by part (a),
$$ c^n - 1 \ge n(c-1) \implies (b^{1/n})^n - 1 = b-1 \ge n(b^{1/n}-1),$$
which is the desired result.
For part (c), you should think about how to use the previous parts to get the result that you want—this may be using "proof by psychology" a bit, but Rudin starts by telling you that this is an outline for a proof, which means that he is trying to hold your hand a little and show you the steps in a reasonable order.  This implies that the later steps should follow from the earlier steps.  In this case:

  Applying part (b) directly to the given inequality: $$ n > \frac{b-1}{t-1} \ge \frac{n(b^{1/n}-1)}{t-1}. $$ Canceling a factor of $\frac{1}{n}(t-1)$ on both sides (i.e. multiplying both sides by $\frac{1}{n}(t-1)$, this becomes $$t-1 > b^{1/n}-1 \implies t > b^{1/n},$$ as desired.

Alternatively, the first thing which came to my mind, which uses part (a) (rather than part (b)), but relies on the monotonicity of the function $x \mapsto x^n$ (which requires some further argument):

 Since $t > 1$, we have that $t - 1 > 0$, so multiply both sides of the given inequality by $t-1$ to obtain $$ n > \frac{b-1}{t-1} \implies n(t-1) > b-1.$$ Applying part (a) to $t$, this implies that $$ t^n - 1 \ge n(t-1) > b-1 \implies t^n > b \implies t > b^{1/n}, $$ where the last implication takes advantage of the fact that the function $x \mapsto x^n$ is an increasing function on the interval $(1,\infty)$.

As an added bit of advice regarding "proof by psychology", if you think that you have come up with counter-examples, you should check your work.  While errors do occur, Rudin's text has been around for a long time and is used by a large number of institutions.  A nasty mistake like the one you purport to have found would likely have been corrected a long time ago.  In your examples, have you really checked all of the hypotheses?  Namely, if $b=144$, $t=10$, and $n = 1/2$, have you checked that:


*

*$b > 1$ and $t > 1$?

*$n$ is a positive integer?  ($\color{red}{\text{Oops!}}$)

*$n > (b-1)/(t-1)$? ($\color{red}{\text{Oops, again!}}$)


What about $b=10$, $t=2$, and $n=3$?  Which hypothesis (or hypotheses) fail(s) here?
