# Find $x$ from $3^x\cdot x^3 = 1$

I saw a question on internet, tried to solve but I can't:

$$3^x\cdot x^3 = 1$$

I get $\ln$ function and made some equalization and I reached that:

$$-x\cdot\ln(3) = 3\cdot\ln(x)$$

I could not continue after that. How does it solve?

Wolfram|Alpha says nearly $x\approx 0.757$.

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Wow. So many people worked on improving this post. Can we please just stop for two minutes? – Asaf Karagila Dec 27 '12 at 18:46
@Amzotti: in $LaTeX$ the common functions get a backslash so they are not italicized. So $\ln$ comes from \ln. Also there is \approx for $\approx$ – Ross Millikan Dec 27 '12 at 18:47

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 $0.7576969788...$

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There are actually three principal solutions. The cube root of $3$ has one real solution and two complex solutions. – Fly by Night Dec 28 '12 at 0:58
@FlybyNight: Not sure what you mean by "principal" solutions. The function $3^x x^3 = e^{x\ln 3} x^3$ is single-valued. Because $W$ is multi-valued, there are actually infinitely many solutions (e.g., $-7.0814 + 11.3443 i$); arguably the only special one is the real one. – mjqxxxx Dec 28 '12 at 3:37

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 $(\frac13,3)$

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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.

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