Solve the equation $x^x=10^9$.

Question

The main question was to solve $x\log_{10}{x}=9$. I reduced it to this equation. This is $x$ Degree equation. How to solve this? I know this can be solved by newton's method. But I am not getting how to apply it.

Answer 1

As Arkamis proposes, you need Lambert's W function.

Then $x^x=10^9 \to x=e^{W(\ln 10^9)}=9.2950869003762184...$

Answer 2

To solve numerically just follow the standard newton raphson method steps:

$$ y =x^x$$

$$y ' = x^x ( \ln x +1 )$$

$$ x \leftarrow x - \frac{y-10^9}{y'} = x - \frac{1 - 10^9 x^{-x}}{\ln x + 1} $$

Answer 3

If you use Newton's method, you are looking for a zero of $x^x-10^9$ by considering $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ and in this case $f(x)=x^x-10^9$ with $f'(x)=x^x(\log_e(x)+1)$.

You need to decide a starting point and obviously $9^9 \lt 10^9 \lt 10^{10}$ so $9$ or $10$ or something between them would do. For example

x_n         f(x_n)      f'(x_n)
9           -612579511  1238670309
9.49454605  908419550.2 6203732894
9.34811493  186970498.1 3840056939
9.29942542  14110834.53 3275529942
9.29511747  98716.84972 3229808064
9.29508690  4.922117114 3229485986

or

x_n         f(x_n)      f'(x_n)
10          9000000000  33025850930
9.72748620  3080960913  13364965418
9.49696104  923461012.5 6253117619
9.34928094  191456569.3 3854718766
9.29961284  14724904.42 3277533809
9.29512016  107408.813  3229836424
9.29508690  5.826986194 3229485989

Answer 4

Your best bet, I'd say, is to recast the problem as looking for a zero of the function

$$f(x)=x\ln x-9\ln10$$

The Newton-Raphson iteration, $x\to x-(f(x)/f'(x))$ is

$$x\to{x+9\ln10\over1+\ln x}$$

Since $9\ln9\lt9\ln10\lt10\ln10$, we know the zero is between $9$ and $10$, so it's sensible to split the difference and start at $x=9.5$. Then you get

$$\begin{align} 9.5&\to{9.5+9\ln10\over1+\ln9.5}\approx9.29577\\ 9.29577&\to{9.29577+9\ln10\over1+\ln9.29577}\approx9.2950869\\ \end{align}$$

etc.