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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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    What do you mean by $\mathbb R^*_+$? – Asaf Karagila Nov 24 '11 at 19:39
  • $\ln x=x$ has no real solutions. Draw a sketch. – JP McCarthy Nov 24 '11 at 19:42
  • Probably $(0, \infty)$. – N. S. Nov 24 '11 at 19:43
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    For $\ln x=x$, answer is that there is no solution, because $\ln x <x$ for all $x$. For proof, calculus is easiest. But if you trust your graphing software, put in the same picture a graph of $y=x$ and a graph of $y=\ln x$. They don't meet. For $\tan x=x$, graphing software will persuade you there are infinitely many solutions. There is the obvious $x=0$. For the others, you will need a numerical method to find good approximations. (If $x$ is a solution, so is $-x$.) For example, there is a solution a bit less than $3\pi/2$. – André Nicolas Nov 24 '11 at 19:51
  • Related: http://math.stackexchange.com/questions/18718/solution-of-tanx-x – leonbloy Nov 24 '11 at 20:19
  • Related: https://math.stackexchange.com/questions/442332/express-solutions-of-equation-tan-x-x-in-closed-form/4399971?noredirect=1#comment9207638_4399971 – IV_ Mar 13 '22 at 13:13

2 Answers2

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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.

Demophilus
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Eric Naslund
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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.

Michael Hardy
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