Having trouble with the following proof.
Given $b > 1, c > 0$, prove that $ \exists \; x$ s.t. $b^{x} < c$.
We can't use $log$, and I have already shown that $b^{x} > c$ by using the formula $b^{n} - 1 = (b-1)(b^{n-1}+...+b+1)$.
I've been thinking of ways to manipulate the above formula, but having no luck. I also thought about trying to use an epsilon delta limit proof, but in the end I would still need to use logs. Any hints or ideas moving forward?
It is not clear what tools are allowed, so I will guess. Let $b=1+t$ where $t$ is positive. Then any positive integer $n$, by the Bernoulli Inequality, we have $(1+t)^n \ge 1+nt\gt nt$.
It follows that $$b^{-n}=\frac{1}{(1+t)^n}\lt \frac{1}{nt}.$$ For suitable $n$, we have $\frac{1}{nt}\lt c$.