Solving $2^x=x^3$ Algebraically

Question

How can I solve $2^x=x^3$ algebraically?

I could take $\log_2(\cdot)$ on both sides, but I'd still be stuck.

Answer 1

As Ahmed S. Attaalla commented, the only analytic solutions are given in terms of Lambert function, namely $$x_1=-\frac{3 W_0\left(-\frac{\log (2)}{3}\right)}{\log (2)}\qquad \text{and} \qquad x_2=-\frac{3 W_{-1}\left(-\frac{\log (2)}{3}\right)}{\log (2)}$$ and you can evaluate them using the series expansion given in the Wikipedia page.

If you do not want or cannot use Lambert function, only numerical methods would do the job. Start graphing your function and you will notice that the roots are around $1$ and $10$. Using them a strating guesses, use Newton method to get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.00000 \\ 1 & 1.61969 \\ 2 & 1.41482 \\ 3 & 1.37493 \\ 4 & 1.37347 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 10.0000 \\ 1 & 9.94143 \\ 2 & 9.93954 \end{array} \right)$$ which are the solutions for six significant figures.

Answer 2

This equation follows the general equation $$a^x = x^{b}$$ and can be solved, for $a \neq 1$, as follows: \begin{align} a^{x} &= x^{b} \\ e^{x \, \ln a} &= x^{b} \\ x^{b} \, e^{- x \, \ln a} &= 1 \\ x \, e^{- x \, \ln a/b} &= 1 \\ - \frac{x \, \ln a}{b} \, e^{- x \, \ln a/b} &= - \frac{\ln a}{b} \\ u \, e^{u} &= - \frac{\ln a}{b} \, \hspace{5mm} u = - \frac{x \, \ln a}{b} \\ u = - \frac{x \ln a}{b} &= W_{0}\left(- \frac{\ln a}{b}\right) \\ x &= - \frac{b}{\ln a} \, W_{0}\left(- \frac{\ln a}{b}\right), \end{align} where $W_{0}(x)$ is the primary branch of the Lambert W-function which is defined as the solution to the equation $W(x) \, e^{W(x)} = x$. Note that in the case $a=1$ the solution is $x = e^{2 \pi i/b}$.

For the problem proposed it is determined that: $$2^{x} = x^{3} \, \text{ leads to } \, x = - \frac{3}{\ln 2} \, W\left(- \frac{\ln 2}{3}\right).$$