$${\dfrac{x-5}{x} = \left(\dfrac{5}{x}\right)^\sqrt{2x+2}}$$

So at the first glance, I thought maybe this was a normal equation and that can be easily solved using logarithm. I was wrong... (after hours of trying).

There are two (possible) ideas which I think possible to solve this problem:

  1. Try to solve $x$ for which, $$0 = \frac {d}{dx} | lhs - rhs |$$I was thinking that the functional graph must be at the minimum if $x$ is the right answer for the equation.
  2. Brute-force all possible real numbers (using computer programming to do this job).

However, I'm also thinking that it must have other ways to solve for $x$ which I have no skills/knowledge whatsoever in order to solve the equation.

This is not my homework, it's just random challenge that popped out while I'm surfing the net!

  • $\begingroup$ What is the source of this interesting problem ? $\endgroup$ – user399078 Feb 18 '17 at 14:56
  • $\begingroup$ @Nirbhay, sincerely it was coming from a Facebook post, some other guy asking the same question on how to solve the problem. $\endgroup$ – Mohd Shahril Feb 18 '17 at 14:57

I guess that only numerical methods could solve the problem and Newton method would probably be the simplest to use.

Starting from a "reasonable" guess $x_0$, Newton mpethod will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, consider that we are looking for the zero of function $$f(x)=\frac{x-5}{x}- \left(\frac{5}{x}\right)^{\sqrt{2 x+2}}$$ I let you the pleasure of finding the derivative. Looking at the plot of $f(x)$, you can see that the $x$ intercept is close to $7$ which would be a perfect guess.

But, let us be lazy and use $x_0=5$ for which $f(5)=-1$. The method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 5 \\ 1 & 6.120046189 \\ 2 & 6.755330011 \\ 3 & 6.902818440 \\ 4 & 6.909062337 \\ 5 & 6.909072846 \end{array} \right)$$ You could do this with Excel computing the derivative numerically.

  • $\begingroup$ thanks for the answer! However, to differentiate f(x) is just pure hard, so instead of Newton's method, I used Secant method and it still gave me the correct answer!! Here is my code on finding approximation to N using Secant method: ideone.com/aU5zWa $\endgroup$ – Mohd Shahril Feb 18 '17 at 17:08
  • $\begingroup$ I learned a new math term today which is "numerical method". I didn't know about that before. Ohh what a good day for me. Thanks again!! $\endgroup$ – Mohd Shahril Feb 18 '17 at 17:08
  • 1
    $\begingroup$ You are very welcome ! In fact, most of equations do not show analytical solutions and require numerical methods. I already spent (enjoying a lot) 57 years in this area. I need to precise that I am not a mathematician but a "numerical analyst". Cheers. $\endgroup$ – Claude Leibovici Feb 18 '17 at 17:13

If you set $5/x=t$, then $$ 2x+2=\frac{10}{t}+2=\frac{10+2t}{t} $$ so the equation can be written $$ (1-t)^{\sqrt{t}}=t^{\sqrt{10+2t}} $$ The limitation $0\le t\le 1$ is implicit in the original equation.

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The left-hand side describes a decreasing function, the right-hand side an increasing function, so the solution is unique. It exists by comparing the values at $0$ and $1$.


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