The illusionist

Take logarithms and write $3\ln(x)=x$ . Now LHS>RHS when $x=3$ but RHS>LHS when $x=1$ and $x=e^2$ . Thus there are $\boxed{2}$ solutions since the graph of $\ln(x)$ is concave.

is it possible to generalize it for $e^x=x^n$ ?

IT SEEMS ANSWER IS 1, BUT NO. IF U SEE THE EQUATION AS e^x/x^3 = 1 and draw its graph by differentiating and checking the nature of graph , the answer is 2

graph of $x^3$ is in I and III quadant. While graoh of $e^x$ is in I and II quadrant . So the common solution will be in I quadrant only. Now graph of $e^x$ starts from $(0,1)$ and graph of $x^3$ starts from $(0,0)$ in I quadrant . Initially for small region slope of $e^x$ is less than $x^3$ as slope of $x^3$ increases very rapidly in square form. So they will intersect in that small region. Now as the x-coordinate increases slope of $x^3$ will become less than $e^x$ . So again they will intersect. Hence there are two soln.