$$x * 0.98^{\sqrt x / 321868} = 9.46 * 10^8$$

I mean is there any way to do this algebraically? There's a single variable, but I don't know of any way to manipulate it such that I can get x on one side -- or even get it out of the exponent, since I can't take the log base 0.98 of the right hand side over a variable...

  • $\begingroup$ You can't do it algebraically. $\endgroup$ Jun 6 '16 at 1:25
  • $\begingroup$ You're right. Is there even a solution? Wolfram $\endgroup$
    – rb612
    Jun 6 '16 at 1:27
  • $\begingroup$ Lol it timed out...even with extended computing time $\endgroup$
    – James
    Jun 6 '16 at 1:38
  • 1
    $\begingroup$ I think there is a solution. I wrote a program to plug in one x and get the other. I find that 9.4783 * 10^8 works. Of course this is limited precision, and it's not what OP asked for. $\endgroup$ Jun 6 '16 at 1:52

Any equation which can write as $$A+Bx+C\log(D+Ex)=0$$ shows solutions which can be expressed in terms of Lambert function.

For the case of $$x\, a^{b \sqrt{x}}=c$$ the solution is given by $$x=\frac{4 W\left(\pm\frac{1}{2} b \sqrt{c} \log (a)\right)^2}{b^2 \log ^2(a)}$$ where $W(z)$ denotes Lambert function.

For your values of $a,b,c$, this will give, as roots, $x_1\approx 9.4783\times 10^8$ and $x_2 \approx 8.5147\times 10^{16}$.

In your specific case, the value of the argument is quite small $(z\approx -0.000965267)$ and you can use the approximation around $z=0$ given in the Wikipedia page $$W(z)=z-z^2+\frac{3 z^3}{2}-\frac{8 z^4}{3}+\frac{125 z^5}{24}+O\left(z^6\right)$$

From a practical point of view, since you handle very large numbers, I would suggest, for a better scaling, some preliminary change of variable such as $$\frac{\sqrt{x}}{321868}=y\implies x=103599009424 y^2$$ which would reduce the equation to $$y^2 \times \left(\frac{49}{50}\right)^y=\frac{59125000}{6474938089}$$ and, taking logarithms, consider the plot of function $$f(y)=2\log(y)+y \log\left(\frac{49}{50}\right)-\log\left(\frac{59125000}{6474938089}\right)$$

$f'(y)$ cancels at $y_*=\frac{2}{\log \left(\frac{50}{49}\right)}\approx 98.9966$ and $f(y_*)\approx 11.8862$. $f''(y)$ being negative for all $y$, $y_*$ corresponds to a maximum and two roots are expected. Graphing the function to locate more or less the roots, then Newton method will lead to solutions $y_1\approx 0.0956505$ and $y_2\approx 906.582$. From here, go back to the solutions for $x$.

  • $\begingroup$ Woah - cool solution $\endgroup$
    – James
    Jun 6 '16 at 4:43
  • $\begingroup$ @James. This is a fantastic function with a lot of applications. Search on this site. You will find plenty of problems using Lambert function. $\endgroup$ Jun 6 '16 at 4:45

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