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I am trying to figure out how to solve for $x$ in the equation $y = (x + 25)^x$ for any known $y$.

For example, I know that in the example $2758547353515625 = (x + 25)^x$, the value of $x$ is 10.

Is there a way to solve for $x$ for any known $y$?

yDivide
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  • integer solution? – 추민서 Sep 24 '20 at 03:15
  • It may be worth noting that I derived the simplest form of this equation from the original equation of $y = \frac{(x + 25)^{x / 5}}{b}$ where both values $y$ and $b$ are known values. – yDivide Sep 24 '20 at 03:15
  • @추민서 The final solution would be rounded down to the nearest integer. – yDivide Sep 24 '20 at 03:18
  • Use Newton's method (https://en.m.wikipedia.org/wiki/Newton's_method) to approximate a solution to as much precision as you want. – Jason Sep 24 '20 at 03:24

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Making the problem more general, you want to find the zero of function $$f(x)=(x+a)^x-y$$

I shall assume $a>0$ and suppose that $y$ could be a huge positive number.

$f(x)$ varies very quickly and this is not good for any root-finding algorithm.

At the opposite, consider instead $$g(x)=x \log(x+a)-\log(y)$$ which is much more linear.

I am very lazy and I shall expand as series to have $$g(x)=x \log (a)+O\left(x^2\right)-\log(y)$$ which gives as an estimate $$x_0=\frac{\log (y)}{\log (a)}$$ which will allow to start Newton method.

Let us take the case of $a=25$ and $y=1234567898765432123456789$. The iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 17.233582865360804139 \\ 1 & 15.056838056149540045 \\ 2 & 15.034341020407527174 \\ 3 & 15.034338495863187264 \\ 4 & 15.034338495863155457 \end{array} \right)$$

Claude Leibovici
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