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I'm not able to solve $$({\frac{1}{2}})^p + ({\frac{1}{3}})^p + ({\frac{1}{7}})^p - 1 = 0.$$

If you put values of $p$ (like $\frac{1}{2}$ or 2) back in the equation it doesn't satisfy! So please check your values also.

What I found is:

If $p = 1$ then the equation becomes $ - \frac{1}{42}$. That means if $p >1$ then values of all the fractions will decrease more and the equation will become more negative. So surely $p < 1$. Similarly putting $p=0$, the equation becomes $-1$ i. e. < $ - \frac{1}{42}$. Hence $ 0 < p < 1. $

Now I'm wondering if there is certain generic techniques to evaluate $p$ ? [ ie. without guessing technique like " Lets assume $p =\frac14(1\pm\sqrt{13}$) " ]

martini
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Arnab Dutta
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2 Answers2

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Not a solution. Just an existence proof:

Firstly $f(p) =\left(\frac{1}{2}\right)^p + \left(\frac{1}{3}\right)^p + \left(\frac{1}{7}\right)^p -1$ is monotone decreasing in p. Next, it is continuous in p.

Now we have $f(0) = 2$ and $f(1) = -1/42$. By the Intermediate value theorem, there is a $0 < \hat{p} < 1$ such that $f(\hat{p}) = 0$. Mvcouwen has verified this via Wolfram Alpha. Moreover as it is monotone decreasing, there are no more solutions in $\mathbb{R}$.

Gautam Shenoy
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First of all, when $p=0$ you should get 2 since $x^0=1\not=0$. As for the solution, note that the left hand side is a monotonically decreasing function so we expect it to have only one root. i.e. only one solution for $p$. As for what that solution is, I've got no clue. You could certainly use a numeric method to find a good approximation to it though.

john
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