Solve the equations $\tan x=x$ and $\ln x=x$ Is it possible to solve the equations $\tan x=x$ and $\ln x=x$ in their respective domains of definition ?
 A: First one: The function $f(x)=e^x-x$ is always positive.  Notice that its second derivative is $e^x$, which is always positive, so we know its concavity.  Looking at $f^{'}(x)=0$, we see that $x=0$ is a solution, so that $f(0)$ is a global minimum.  But $f(0)=1>0$.
Second one:  $\tan(x)-x$ will have infinitely many zeros since $\tan$ is unbounded on any interval of length $\pi$.  Specifically $0$ is one such solution.
A: If you superimpose the graphs of $y=x$ and $y=\tan x$ (do this in radians) you see infinitely many points of intersection.  Their $x$-coordinates are the solutions.  And you immediately see that one of those is $x=0$.  (Notice that the slope of $y=\tan x$ is $1$ at $x=0$ but then it gets steeper if you go in either direction from there.  That implies it can't have other solutions close to $x=0$.  "Close" in this case could be taken to mean in the same period of the tangent function, i.e. between $-\pi/2$ and $\pi/2$.)
The equation $x=\ln x$ has no real solutions.  That can also be quickly seen by superimposing their graphs, and notice that the slope of $y=\ln x$ is $1$ at the point where it crosses the $x$-axis.
