Original Math Proof - When does $y=b^x$ intersect the line $y=x$ if $b>1$?
When does $y=b^x$ intersect the line $y=x$ if $b>1$? by Joseph Kim at josephkimlife.blogspot.com February 2019 $\text{If } y=b^x \text{ and } y=x \text{, then }x=b^x \text{ and }b = x^{1/x} \text{. We would like this to have only one}$ $\text{solution for } b>1 \text{. Now we will prove that }b \leq e^{1/e} \text{ :}$ \begin{align*} \lim_{n\rightarrow \infty}{\left(1+\frac{1}{n}\right)}^n &= e \\ \lim_{m\rightarrow 0}{\left(1+m\right)}^{1/m} &= e \\ (1+m)^{1/m} &\leq e \\ \text{(with equality only } &\text{when m=0)} \\ 1+m &\leq e^m \\ 1+\frac{(x-e)}{e} &\leq e^{(x-e)/e} \\ \frac{x}{e} &\leq e^{x/e-1} \\ x &\leq e^{x/e} \\ x^{1/x} &\leq e^{1/e} \\ b &\leq e^{1/e} \\ \end{align*} $\text{Note that, in the above proof, we must have } b\geq0 \text{, but we are already restricted to } b>1,$ $\text{so this condition is irrelevant. Under this restriction, } e^{1/e} \text{ is an absolut
