I think I'm not following this discussion very well. Just stabbing around with a Haskell window gave me x = 1.14331628080... in just a couple of minutes. The function isn't doing anything exotic in this range, so I would think that a Newton iteration would converge on the answer very fast. What does "abysmal" convergence mean in this context? On Sun, Oct 21, 2018 at 12:43 PM Henry Baker <hbaker1@pipeline.com> wrote:

Substituting x=log(y), we get

y^log(a)-y^log(b) = c

i.e., y^log(50)-y^log(49)=2

But this "polynomial" may not be any easier to solve.

We could approximate log(a)/log(b) with a continued fraction and solve a trinomial, but it might have a quite high degree.

For example, if we let y = z^49.33508819661897 in y^log(50)-y^log(49)=2,

we get z^193-z^192=2, which doesn't factor over Q, and Maxima can't solve it.

At 07:26 AM 10/21/2018, Bill Gosper wrote:

...

I usually suggest the techniques in gosper.org/newsrope.pdf , but everything I've tried for 50^x - 49^x = 2 converges abysmally.

—rwg

On 2018-10-19 12:41, Henry Bakerr wrote:

...

I can solve this equation with Newton's iterations, but I was curious if there might be an elegant closed form solution.

I tried fiddling with hyperbolic trig functions, but nothing seemed any simpler.

real a>b>0,c>0

E.g., a ~ 50, b ~ 49, c ~ 2 => x ~ 1.1438929

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