Find $x$ from $3^x\cdot x^3 = 1$ I saw a question on internet, tried to solve but I can't:
\begin{equation}
  3^x\cdot x^3 = 1
\end{equation}
I get $\ln$ function and made some equalization and I reached that:
\begin{equation}
  -x\cdot\ln(3) = 3\cdot\ln(x)
\end{equation}
I could not continue after that. How does it solve?
Wolfram|Alpha says nearly $x\approx 0.757$.
I searched on Google and on that site but I could not found anything (I guess about the mathematical sign).
 A: I don't believe the equation
$$-x\ln 3=3\ln x\iff 3\ln x+x\ln 3=0$$
can be solved by radicals, the solution probably no closed form . You can do the following however:
The function $$f(x)=3\ln x+x\ln 3$$ is strictly increasing and so it has at most one root. In addition, $f(\frac13)=-3\ln 3+\frac13\ln 3<0$ while $f(3)=6\ln 3>0$ and because $f$ is continuous, $f$ has a root in $(\frac13,3)$ (Intermediate Value Theorem). Therefore, there is only one solution of your equation and is in $(\frac13,3)$
A: Try Newton's method on $f(x) = x^3 3^x - 1$.  Begin with a guess of $x_0=1$:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
where $f'(x) = x^2 3^x (3 + x \log{3})$
As an example, $x_1 = 1 - \frac{3-1}{3 (3 + \log{3})} \approx 0.837343$
Repeat until you get the desired number of significant figures.
A: Equations like this can be solved in terms of the Lambert W function, which is the inverse of $f(z)=ze^z$ (and generally they don't have closed-form solutions in more elementary functions).  Specifically, consider the equation $y \ln y = e^{\ln y}\ln y = k$.  This inverts to $\ln y=W(k),$ or $y=e^{W(k)}$.  In your case, you have
$$
\frac{\ln 3}{3}=-\frac{1}{x}\ln x=\frac{1}{x}\ln\frac{1}{x},
$$
implying that
$$
\frac{1}{x}=\exp\left[W\left(\frac{\ln 3}{3}\right)\right]=e^{W(\ln \sqrt[3]{3})},
$$
or
$$
x=e^{-W(\ln \sqrt[3]{3})}.
$$
This correctly evaluates to $0.7576969788...$
