How do i solve $x^2/3=2/3^x$

Question

How is this done?

$${x^2\over3}={2\over3^x}$$

I'm out of ideas.

It seems simple but it is quite not so the case in my perception of it.

Answer 1

As already said in comments and answers, Lambert function would allow to provide the exact closed form solution and Newton method will easily do the job.

We can also get the solution using Taylor series. Consider the function $$f(x)={x^2\over3}-{2\over3^x}$$ By inspection, we can say that the solution is close to $1$ since $f(1)=-\frac 13$ and $f(2)=\frac {10}9$. For conveniency, rewrite the function as $$f(x)={x^2\over3}-2e^{-x\log(3)}$$ and build Taylor series at $x=1$. So, we obtain $$f(x)=-\frac{1}{3}+\frac{2}{3} (1+\log (3))(x-1)+\frac{1}{3} \left(1-\log ^2(3)\right)(x-1)^2+\cdots$$ Solving the quadratic, we obtain the solution which is given by $$x=\frac{\log ^2(3)+\log (3)-\sqrt{2+\log (9)}}{\log ^2(3)-1}\approx 1.24112$$ which is not too bad as an approximation.

We could repeat the process building now a Taylor series at $x=\frac 54$; this will give $$f(x)=\left(\frac{25}{48}-\frac{2}{3 \sqrt[4]{3}}\right)+ \left(\frac{5}{6}+\frac{2 \log (3)}{3 \sqrt[4]{3}}\right)\left(x-\frac{5}{4}\right)+\frac{\left(3-3^{3/4} \log ^2(3)\right)}{9} \left(x-\frac{5}{4}\right)^2$$ Solving the quadratic woud give $x\approx 1.23973$ which is better (but significantly more complex to get).

Please notice that if the series were limited to first order, the calculation will be equivalent to one step of Newton method.

Edit

Just for your curiosity, this problem could also be approached using Pade approximants (may be, you never heard about them but you will do at a time). These approximate $$f(x)=\frac{P_m(x)}{Q_n(x)}$$ where $P$ and $Q$ are polynomials.

In the case of the function we are looking for, the simplest approximation built at $x=1$ is $$f(x)=\frac{\frac{1}{6} (x-1) (5+3 \log (3))-\frac{1}{3}}{\frac{1}{2} (x-1) (\log (3)-1)+1}$$ So, for the solution, we just need to look for the zero of the numerator which is given by $$x=\frac{7+3 \log (3)}{5+3 \log (3)}\approx 1.24108$$ which is nicer expression and much easier to evaluate.

Answer 2

The only real solution is $$\dfrac{2}{\ln(3)} W(\sqrt{6} \ln(3)/2) $$ where $W$ is the Lambert W function. Numerical value approximately $1.239726175$.

If you don't like Lambert W, use Newton's method for numerical approximations.