The roots of equation $x3^x=1$ are I have to find roots of equation
 $x3^x=1$
A.Infinitely many roots
B.$2$ roots
C.$1$ root
D. No roots\
How do i start? Thanks
 A: Hint: 
$$3^x = \frac{1}{x}$$ Has only $1$ solution.
A: For a simple answer plot the graphs of $y_1=3^x$ and $y_2=\frac{1}{x}$, (that are elementary), and see that these graph intersects only at a point $x_0$ such that $0<x_0<1$  because:
1) $3^0=1$ and $ \frac{1}{x} \to +\infty$ for $x \to 0^+$
2) $y_1(1)=3^1=3>y_2(1)=\frac{1}{1}=1$
3) the two functions are continuous in $(0,1]$.
4) for $x>0$ $y_1$ is monotonic increasing and $y_2$ is monotonic decreasing and for $x<0$:  $y_1>0$ and $y_2<0$.
If you want the value of $x_0$ this cannot be done with elementary functions. You can use the Lambert $W$ function that is defined as:
$$
W(xe^x)=x
$$
so, from
$$
x3^x=1 \iff xe^{x\ln 3}=1
$$
using $x\ln 3=t$ we find:
$$
te^t=\ln 3\quad \Rightarrow \quad t=W(\ln 3)
$$
and
$$
x_0=\frac{W(\ln 3)}{\ln 3}
$$
A: Define the function $f(x)= x3^x-1= x\exp(ln(3)x)-1$ and study the variation of this function on $\mathbb{R}$. 
A: We can solve this by Newton raphson method.
$x3^x=1$
By solving i got $x$ approximately equal to 0.5478
The given problem have only one solution( by graphical method )
