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Is it possible to solve the equations $\tan x=x$ and $\ln x=x$ in their respective domains of definition ?

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    Related: http://math.stackexchange.com/questions/18718/solution-of-tanx-x2011-11-24

2 Answers 2

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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.

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    @ChrisTaylor: Oops that is what I mea$n$t! Thanks.2011-11-24
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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.