Given $a,b,c$ solve $1=x^a + x^b - x^c$ for $x$

Asked Mar 11 '21 at 18:19

Active Mar 11 '21 at 18:28

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If $a,b,c$ are constants, is it possible to isolate $x$ in the following equation?

$$1 = x^a + x^b - x^c$$

If so, how do I achieve this?

edited Mar 11 '21 at 18:28

asked Mar 11 '21 at 18:19

fuseinabowl

Do you mean, solve for $x$? If so, there are no general methods. – lhf Mar 11 '21 at 18:22
I think he wants a general formula for $x$. – Rounak Sarkar Mar 11 '21 at 18:25
@lhf yes. Thanks - if you post as an answer I will accept. – fuseinabowl Mar 11 '21 at 18:27
Are $a,b,c$ positive? It's not hard to find $x$ numerically, given $a,b,c$. The only slightly tricky thing is avoiding the trivial solution $x=1$. – PM 2Ring Mar 11 '21 at 18:29
Unfortunately not, in my use-case both $b$ and $c$ are negative – fuseinabowl Mar 11 '21 at 18:31
1

What about the sign of a ? Besides, are $a,b,c$ integer or real ? – Jean Marie Mar 11 '21 at 18:33
Can you give me a typical set of $a,b,c$ so I can test how well my strategy works? – PM 2Ring Mar 11 '21 at 18:34
$a$ is always positive. All of $a,b,c,x$ are real. I will collect a set of $a,b,c,x$ for you. – fuseinabowl Mar 11 '21 at 18:38
@PM2Ring here's a sample: $a=0.1, b=-18, c=-17$. This should give $x=e$ if I've done my stuff right. – fuseinabowl Mar 11 '21 at 18:43
$a = 0.1, b=-14.42419, c=-13.42419$, giving $x=e^{0.1}$ – fuseinabowl Mar 11 '21 at 18:45
I've just calculated these forwards and got a value of $1.105170$ for both of them (they should give $1$). I'm getting my values from a program I am currently writing so I must have made a mistake somewhere, I wouldn't expect a 10% inaccuracy here. – fuseinabowl Mar 11 '21 at 18:51
Thanks. As I suspected, the $x=1$ solution causes problems for a simple Newton's Method approach. I'll have to think about this... – PM 2Ring Mar 11 '21 at 18:54
Ok. For (0.1, -18, -17) I get $x=0.14024358282171$. Here'a a Sage / Python script using Newton's method. It needs a good initial approximation for $x$. To run it in plain Python, change all ^ to **. – PM 2Ring Mar 11 '21 at 20:33
@PM2Ring This works great! Thank you very much for your time. – fuseinabowl Mar 12 '21 at 15:38
No worries. There might be a good way to produce a good initial approximation, but simple trial & error seems to work ok. ;) – PM 2Ring Mar 12 '21 at 15:40

Given $a,b,c$ solve $1=x^a + x^b - x^c$ for $x$

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