How to solve $x3^x=18$ algebraically

Question

I've been struggling with $x3^x=18$ for a long time. I aware of the existence of the Lambert $W$ function and I know its definition as the inverse of $xe^x$ but I don't know how it's used in Algebra. Numerical methods show that the integer answer is $2$, but I'm looking for all solutions.

Answer 1

Let $f(x)=x3^x$.

Thus, $f$ increases for $x>0$ without derivative of course.

Thus, our equation has one root maximum for $x>0$.

But $2$ is a root and for $x\leq0$ our equation has no roots.

Thus, $2$ is an unique root.

Answer 2

now you will get $$\ln(x)+x\ln(3)=\ln(18)$$ and with $$f(x)=\ln(x)+x\ln(3)-\ln(18)$$ we obtain $$f'(x)=\frac{1}{x}+\ln(3)>0$$

Answer 3

Note that

$$f(x)= x3^x \implies f'(x)= 3^x(x+ln3)>0$$

is strictly increasing and

$$f(-1)=-\frac13 \quad f(3)=81 $$

thus for IVT a $f(x)=0$ unique solution exists in the interval $(-1,3)$ that is $x=2$.

Answer 4

Trying to avoid use of derivatives... Obviously, if a solution $x$ exists, it must satisfy $x\gt 0$ (because $3^x\gt 0$ and $18\gt 0$). On the interval $(0,\infty)$ the function $f(x)=x\cdot 3^x$ is strictly increasing, because $x\lt y$ implies $3^x\lt 3^y$ and then $x\cdot 3^x\lt x\cdot 3^y\lt y\cdot 3^y$.

Thus, if there is a solution, it is unique. We notice that $x=2$ is a solution, and then it must be the only one.