Re: Solving simple equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg122737] Re: Solving simple equations*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 8 Nov 2011 07:16:16 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201111071054.FAA03874@smc.vnet.net>

I think one should note that when using PowerExpand one can get a solution that is only valid for some values of the parameters or even one that is never valid at all. As an example, consider the equation: eq = Sqrt[1/z] - 1/Sqrt[z] + Sqrt[1/a] - 1/Sqrt[a] + a + z == 0 It actually has no solutions over the reals as you can tell with Reduce[eq, z, Reals] False But if you use PowerExpand you will get a "solution": sol = Solve[PowerExpand[eq], z, Reals][[1]] {z->-a} The solution is, of course, wrong. If you use PowerExpand with Assumptions you can see this fact: Solve[PowerExpand[eq, Assumptions -> a >= 0], z, Reals] {{z -> ConditionalExpression[-a, a < 0]}} The answer contradicts the assumption. Similarly: Solve[PowerExpand[eq, Assumptions -> a >= 0], z, Reals] {{z->ConditionalExpression[-a,a<0]}} Andrzej Kozlowski On 7 Nov 2011, at 11:54, Dana DeLouis wrote: >> ...Reduce (& Solve) doesn't work... >> ...Solve[(A*n^a)^b/n == c, n] > > Hi. The only way I found to get a real solution for this particular problem is to use PowerExpand: > > data = {A -> 2, a -> 3, b -> 4, c -> 5.}; > > equ = PowerExpand[(A*n^a)^b/n == c] ; > > Solve[equ, n] > > Which gives: > > n -> (c/A^b) ^ (1/(a*b - 1)) > > % /. data > 0.8996576447849666 > > Check with Solve using the same data... > > (A*n^a)^b/n == c /. data > > 16*n^11 == 5. > > NSolve[%, n, Reals] > > n -> 0.8996576447849665 > > = = = = = = = = = = > HTH > Dana DeLouis > = = = = = = = = = = > > > > On Nov 5, 4:57 am, Mathieu <mat... at gmail.com> wrote: >> Mathematica seems to struggle with very simple equations: >> >> Solve[(A*n^a)^b/n == c, n] >> Solve::nsmet: This system cannot be solved with the methods available >> to Solve >> >> Reduce doesn't work either. Even if I add assumptions: >> Assuming[A > 0 && 1 > a > 0 && 1 > b > 0 && c > 0 && n > 0, Solve[(A >> n^a)^b/n == c, n]] >> Solve::nsmet: This system cannot be solved with the methods available >> to Solve. >> >> >> Is there a way for Mathematica to solve these equations? >> >> Many thanks, >> Mathieu > > >

**References**:**Re: Solving simple equations***From:*Dana DeLouis <dana01@me.com>