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Find the all real solutions to $$ x(x-1)^{x-1/x}=1 $$

This question was asked to 10th grade students by a teacher who was preparing questions for the contest mathematics.The teacher said she would share the original answer to the question soon, but I can't wait. I wanted to ask here.

WA could not find the exact answer and gives an approximate solution as $$ x\approx 1.61803398874989... $$ The equation seems really difficult and the way to approach the solution is not clear. There are very few elementary ways to try. Graphically the equation has clearly one real root.

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How do we solve that strange equation?

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    $\begingroup$ The result seems to be the golden ratio. $\endgroup$
    – Gonçalo
    Commented Sep 9, 2023 at 20:08
  • $\begingroup$ Golden Ratio ? Fantastic !! $\endgroup$
    – hardmath
    Commented Sep 9, 2023 at 20:10
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    $\begingroup$ The golden ratio $\phi$ satisfies the equation $\phi^2-\phi-1=0$, hence $\phi-\frac{1}{\phi}=1$ and $\phi^2-\phi=1$, therefore $\phi$ is a solution to the equation $x(x-1)^{x-1/x}=1$: $$ \phi(\phi-1)^{\phi-1/\phi}=\phi(\phi-1)^1=\phi^2-\phi=1. $$ $\endgroup$
    – Gonçalo
    Commented Sep 9, 2023 at 20:32

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I'm not sure if there is an intuitive way to see this other than noting that the golden ratio solves $x^2 - x-1 = 0$ and factoring

$$\begin{align} &x(x-1)^{x-1/x} \\ &=x(x-1)^{(x^2-1-x+x)/x} \\ &=x(x-1)\cdot (x-1)^{\frac{x^2-x-1}{x}} \\ &=(1 + [x^2-x-1])\cdot (x-1)^{\frac{x^2-x-1}{x}} \end{align}$$

Subbing in the larger root of the polynomial gets your answer (subbing in the smaller root is invalid,) but I'm not sure how exactly one derives this without some sort of guess already.

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    $\begingroup$ Can you use $1/\phi = \phi - 1$? $\endgroup$
    – qwr
    Commented Sep 10, 2023 at 9:49

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