How to show $\log x$ and $e^x$ never meet.

Question

I want to show $\log x$ and $e^x$ never meet without having any graphing calculator.

Answer 1

I'm not sure whether you are familiar with calculus, but that would be my approach. I'm also not sure if you're using "log" to mean natural log (that's what I use in my answer) or log base 10. My answer can be adapted to work for log base 10 as well.

First note that $e^1=e$ and $\log(1)=0<e$, so $e^x> \ln(x)$ at $x=1$.

Next, $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}\ln(x)=\frac{1}{x}$. Thus $\frac{d}{dx}e^x >e$ for $x>1$, and $\frac{d}{dx}\log(x) < 1<e$ for $x>1$. So the derivative of $e^x$ is greater than the derivative of $\ln(x)$ for $x>1$. Hence, if $x>1$: $$e^x =\int_1^xe^t \ dt +e^1>\int_1^x \frac{1}{x} \ dx = \ln(x).$$

For $x<1$ we have $e^x>0>\ln(x)$.