Prove (without derivatives) that for any $b\in \mathbb{R}$, $b>1$, that $ b^x >1$ for all $x\in \mathbb{R}$, $x>0$

Question

I am trying to prove (without using any differential calculus) that $b^x > 1$ for $b\in \mathbb{R}$, $b>1$ and $x \in\mathbb{R}$ for $x>0$. I was wondering if anyone had any hints on how to do this.

Answer 1

I assume you know that exponential and the logarithm are strictly increasing functions. $b^x =exp(xln(b))$ since $b>1$, $ln(b)>0$ and $xln(b)>0$ use that exponential is strictly increasing to conclude that $b^x=exp(xln(b))>exp(0)=1$.