Re: Solving Tanh[x]=Tanh[a]Tanh[b x + c]

*To*: mathgroup at smc.vnet.net*Subject*: [mg83545] Re: Solving Tanh[x]=Tanh[a]Tanh[b x + c]*From*: dh <dh at metrohm.ch>*Date*: Thu, 22 Nov 2007 04:58:10 -0500 (EST)*References*: <fhu7h8$79u$1@smc.vnet.net>

Hi Yaroslav, you must prevent the "uphill battle" by e.g. temporarily write Exp[a]Exp[b] as {Exp[a],Exp[b]}, then do what you want and finally eliminate the braces.E.g: ... //.{Exp[a__+b_]->{Exp[a],Exp[b]},Exp[2x]->x,{a_,b_}->a b} hope this helps, Daniel Yaroslav Bulatov wrote: > I'd like to use Mathematica to show that solution of Tanh[x] - Tanh[a] > Tanh[b x + c]=0 can be written as > 1/2 Log (Root[c1 x^(1+b) + c2 x^b + c3 x -1]) for certain coefficients > c1,c2,c3 when b is a positive integer > > Tanh[x] - Tanh[a] Tanh[b x + c]// TrigToExp // Together // Numerator > gives me almost what I need, except now I need to factor out Exp[2x] > as a separate variable. What's the best way of achieving it? Using > syntactic replacement rules like {Exp[a_+b_]->Exp[a]Exp[b],Exp[2x]->x} > seems like an uphill battle against the evaluator which automatically > simplifies Exp expressions > > Yaroslav >