2

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.

math_guy
  • 83
  • Depending on how approximate you want it, note that the RHS is equal to $0.003465...$, which is pretty close to $0$. Additionally, recall that $\ln(1)=0$. Putting these two things together lets you know that one solution must be $1+\epsilon$, $\epsilon>0$. – scoopfaze May 04 '20 at 07:53
  • There are two real solutions, $1.003483864655896182306486425424729153476171355495599541471\ldots$ and $2223.756639530995875729099468820912339080066882151155014582\ldots$. –  May 04 '20 at 08:21

1 Answers1

3

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 $$2242.20\cdots$$