Finding roots of a mathematical equation

Question

If $x^3 + 3^x = 0$ what will be the value of $x$? what will be the easiest way to find it mathematically?

Answer 1

Usually you would solve this numerically, using e.g. Newton's method.

However it can be solved exactly in terms of the Lambert W function $$x=\frac{-3W\left(\frac{\log(3)}{3}\right)}{\log(3)}$$

Answer 2

Put

$$f(x)=x^3+3^x\implies f'(x)=3x^2+3^x\log3>0\implies f(x)\;\;\text{is monotonic ascending}$$

so $\;f(x)=0\;$ has at most one single solution.

Now, we have that $\;f(-1)\cdot f(0)=\left(-\frac43\right)\cdot1<0\;\implies$ the zero of $\;f(x)\;$ is in $\;(-1,0)\;$ . You any want to continue with bisecting this interval or use any other numerical method (Newton, Rapson-Newton, Lambert Function, etc.) to approximate it.

Answer 3

$$x^3 + 3^x = 0 \iff x^3 = -3^x$$

If $x$ is real, $$\implies x = -3^{\frac{x}{3}} \implies3 ^{-\frac{x}{3}} x = -1 \implies xe^{-\frac{x}{3} \ln3} = -1 \implies -\frac{x}{3} \ln(3) e^{-\frac{x}{3} \ln3} = \frac{\ln 3}{3}$$

So $$-\frac{x}{3} \ln 3 = W(\frac{\ln 3}{3})$$

$$x = - \frac{3W(\frac{\ln 3}{3})}{\ln 3 }$$

If x is complex, $$x = -3^{\frac{x}{3}} \omega $$ or $$x = -3^{\frac{x}{3}} \omega^2 $$ where $\omega$ is a primitive root of order $3$.

In the same way, one will get $$x = - \frac{3W(-\frac{1}{6} i (-i \ln 3 + \sqrt 3 \ln 3 ))}{\ln 3 }$$ and $$x = - \frac{3W(\frac{1}{6} i (i \ln 3 + \sqrt 3 \ln 3 ))}{\ln 3 }$$

Where W is Lambert W function and $i$ is unity of imaginery.