Approximating the solution to $2^x=x^{200}$

Question

I'm wondering if there exists some numerical method (maybe using a series) for approximating the solution to an equation like $2^x=x^{200}$. By rewriting both sides of this equation we have $e^{x\ln 2}=e^{200\ln x} $, so the curves intersect when $\frac{\ln x}{x} = \frac{\ln 2}{200} $. I know this cannot be solved exactly but what could one do to find the approximate value of $x$? Thanks in advance.

Answer 1

As $\dfrac{\ln 2}{200}$ is small, one of the roots is close to $x=1$.

We can improve with a few Newton's iterations,

$$x_{n+1}=x_n-\frac{200\ln x_n-x_n\ln 2}{\dfrac{200}{x_n}-\ln 2}.$$

The next approximation is

$$1+\frac{\ln 2}{200-\ln 2}\approx1.00347779\cdots$$

For the other root, which is large, if we consider the inverse $y:=\dfrac1x$, we have

$$y\ln y=-\dfrac{\ln 2}{200}.$$

The LHS has no Taylor development around $0$, but we can approximate the function linearly between $0$ and a small value, say $-\dfrac{\ln2}{200}$.

So

$$y\ln y\approx y\ln\dfrac{\ln2}{200}$$ giving

$$x=\frac1y\approx\frac{\ln\dfrac{\ln2}{200}}{\dfrac{\ln2}{200}}=1634.5\cdots.$$

Then next approximation is $$2242.20\cdots$$