Find all possible values of $x$ if $\ln(x) = \sin(x)$. My Mathematics Teacher gave me the following problem :

Find all the possible values of $x$ if $\ln(x) = \sin(x)$.

I tried graphing both $\ln(x)$ and $\sin(x)$. I found that they intersect at a single point $\approx (2.22, 0.8)$. However, I am unable to find a "mathematical" solution to this problem using the elementary properties of trigonometry and logarithms. I tried some Taylor expansions, but that did not help. Can anybody provide a hint as to what should be done ?
Also, what can be the possible values of $x$ is the problem is generalized as $\sin(x)=\log_b(x) ? $ One simple observation is that $0<x\leq b$ as maximum value of the sine function is 1. How to proceed ?
 A: As said in answers and comments, the solution requires numerical methods.
Since you graphed the functions, you noticed that they intersect close to $x=2$. Then, in order to approximate the solution, you could have used Taylor expansions (as you though about) but building them around $x=2$.
You have $$\log(x)=\log (2)+\frac{x-2}{2}-\frac{1}{8} (x-2)^2+\frac{1}{24}
   (x-2)^3+O\left((x-2)^4\right)$$ $$\sin(x)=\sin (2)+(x-2) \cos (2)-\frac{1}{2} (x-2)^2 \sin (2)-\frac{1}{6} (x-2)^3 \cos
   (2)+O\left((x-2)^4\right)$$
Limited to first order, we then have $$\log(x)-\sin(x)=(\log (2)-\sin (2))+(x-2) \left(\frac{1}{2}-\cos (2)\right)+O\left((x-2)^2\right)$$ leading to the solution $$x_{(1)}=\frac{2 (-1+\log (2)-\sin (2)+2 \cos (2))}{2 \cos (2)-1}\approx 2.23593$$ Limited to second order,we  have $$\log(x)-\sin(x)=((\log (2)-\sin (2))+(x-2) \left(\frac{1}{2}-\cos (2)\right)+(x-2)^2
   \left(\frac{\sin (2)}{2}-\frac{1}{8}\right)+O\left((x-2)^3\right)$$ and solving the quadratic would give $$x_{(2)}\approx 2.21872$$ while the "exact" solution given by Newton method would be $$x\approx 2.21911$$
Edit
May be, it would be nicer to build Taylor expansion around $x=\frac{2\pi}3$. This would give 
$$\log(x)-\sin(x)=\left(\log \left(\frac{2 \pi }{3}\right)-\frac{\sqrt{3}}{2}\right)+\frac{(3+\pi )
  }{2 \pi } \left(x-\frac{2 \pi }{3}\right)+\left(\frac{\sqrt{3}}{4}-\frac{9}{8 \pi
   ^2}\right) \left(x-\frac{2 \pi }{3}\right)^2+O\left(\left(x-\frac{2 \pi
   }{3}\right)^3\right)$$ Limiting to first order, the solution would be 
$$x_{(1)}=\frac{\pi  \left(6+3 \sqrt{3}+2 \pi -6 \log \left(\frac{2 \pi }{3}\right)\right)}{3
   (3+\pi )}\approx 2.22408$$ Using the second order, this would give $$x_{(2)}\approx 2.21901$$
A: I can't think of anything other than a numerical method to find the answer.
However, you can use purely mathematical methods to prove that there is only one answer.
$\ln x < \sin x$ for all $x\le \frac {\pi}2$
$\ln x > \sin x$ for all $x\ge \frac {3\pi}2$
IVT says that there is at least one point in $(\frac {\pi}2, \frac {3\pi}2)$ where $\sin x = \ln x$
Since $\frac {d}{dx} \sin x < 0$ and $\frac {d}{dx} \ln x > 0$ for all $x$ in this inteval there is only one point in the inteval where $\sin x = \ln x$
A: Observe that $\ln$ is an unbounded, strictly increasing function, and that $\sin$ is periodic with range $-1 \le \sin(x) \le 1$.  So the only area where there can be such a solution is in the area $1/e\le x \le e$, the area where $-1 \le \ln(x) \le 1$.
We can then break this area up into three parts: 


*

*$1/e \le x < 1$, where both $\ln(x)$ and $\sin(x)$ are increasing and $\ln(x)<0$ but $\sin(x)>0$, so there cannot possibly be an intersection there,

*$1 < x < \pi/2$, where both are increasing and positive; however, in this area, $\ln(x)$ ranges from $0$ to about $0.45$, and $\sin(x)$ ranges from about $0.82$ to $1$, so there still cannot possibly be an intersection there,

*and finally $\pi/2 < x \le e$, where $\ln(x)$ is increasing and $\sin(x)$ is decreasing; this is where they cross, and since the slopes are now opposite signs, there must be only one.


Once we've done this, we can use standard numerical root-finding techniques to locate the root.  There's no hope in finding an analytic solution.
