I was trying to solve the following problem from a calculus book
Show that the graph of $e^x$ and $\ln x$ do not intersect.
I tried to solve the problem by showing the function \begin{align}\label{1} f(x)=e^x-\ln x \tag{1} \end{align} does not have a zero (which is another problem). So, \begin{align}\label{2} f'(x)=e^x-\frac1x\tag{2} \end{align} In order to find the minimum of (\ref{1}), we should find the zero of (\ref{2}), yet another problem.
If you plot the two functions $y=e^x$ and $y=\ln x$ you'll notice that you can fit the line $y=x$ between both of these. So one approach would be to prove that $e^x>\ln x$ in two steps:
$e^x > x$ for all $x$.
$x > \ln x$ for all $x$.
These two assertions should be easier to attack using your approach of finding the minimum of the difference.
From the series expansion of $e^x$ you can get that for any $x \in \mathbb{R}$, $e^x \ge 1+x \gt x$. Thus the graph of $y = e^x$ lies above the graph of $y = x$. Since $\ln x$ is the inverse function of $e^x$, it's graph is the reflection of the graph of $e^x$ over the line $y = x$. Thus since $e^x$ never intersects $y = x$, it cannot intersect $\ln x$.
Your way leads quickly to a result:
You already received good answers; so, just for your curiosity, may I welcome you in the world of Lambert function !
$$f'(x)=e^x-\frac 1x = 0 \implies x=W(1)=\Omega \approx 0.567143$$ From here, everything becomes quite simple since, at this point $$f(x)=\Omega -\log(\Omega) >0$$
Sooner or later, you will learn that any equation which can write $A+B x+C\log(D+ex)=0$ has solution(s) in terms of Lambert function?
Edit
This is totally off-topic but used for illustration.
Let us try to find the minimum value of $k$ such that the curves $y=e^x$ and $y=k \log(x)$ intersect.
So, we look for the possible zeros of function $$g(x)=e^x-k \log(x)$$ $$g'(x)=e^x-\frac k x=0 \implies x=W(k)$$ Taking into account that $k=W(k)\,e^{W(k)}$, at the minimum, we then have $$g(W(k))=k\left(\frac{1}{W(k)}- \log (W(k))\right)$$ Let $z=W(k)$ to solve again $$g(z)=\frac 1 z -\log(z)=0\implies z=\frac 1 \Omega\implies k_*=\frac{1} \Omega e^{\frac 1 \Omega} \approx 10.2817$$ Since $g''(x) > 0 \,\,\forall k >0$, for any $k > k_*$, $g(x)=0$ will show two roots $0 < x_1 < W(k)$ and $x_2 > W(k)$.
1) $x \in (0,1]$ :
$e^x >0$, $\ln x \le 0$, $\\ $no intersection.
2) $x \in (1,\infty):$
$\star)$ $e^x =1+x+x^2/2!+...> x$, since
$\ln$ is strictly increasing: $\ln e^x > \ln x$ , or
$x > \ln x;$
Combining with $\star)$:
$e^x >x > \ln x $, not Intersecting.
The graphs of $e^x$ and $\log x$ in the first quadrant are symmetric with respect to the line $y=x$ since $\exp$ and $\log$ are inverse functions. Since $e^x\geq x+1$ holds by convexity we have $$ e^x \geq x+1 > x > x-1 \geq \log(x) $$ for any $x>0$.
With $\displaystyle x > 0$, note that $\displaystyle\quad\mathrm{e}^{x} > 1 + x\quad$ and $\displaystyle\quad\ln\left(x\right) \leq x - 1 \implies \bbox[10px,border:1px groove navy]{\large\mathrm{e}^{x} - \ln\left(x\right) > 2}$
