Re: Re: Exponential Equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg29844] Re: [mg29808] Re: Exponential Equations*From*: "J Rockmann" <mtheory at msn.com>*Date*: Fri, 13 Jul 2001 04:19:24 -0400 (EDT)*References*: <9hetrp$6ku$1@smc.vnet.net> <200107110526.BAA22228@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

----- Original Message ----- From: "Erich Mueller" <emuelle1 at uiuc.edu> To: mathgroup at smc.vnet.net Subject: [mg29844] [mg29808] Re: Exponential Equations > > The exponential function is periodic in the complex plane. The equation > that you give has an infinite number of solutions, x=-1+(n/7) Pi I, for > integer n, so you should not be surprised to find multiple solutions. > > Mathematica's equation solving algorithm is made for solving > algebraic equations, so it solves your problem by factoring a seventh > order polynomial in Exp[x] -- hence the seven solutions. > > only finds 7 of the solutions. > > On Thu, 28 Jun 2001, Shippee, Steve wrote: > > > If this is too basic of a question for the "LIST", please respond to me > > directly via email at shippee at jcs.mil and thanks in advance for any > > assistance provided. I'd be happy to provide you a "notebook", too, if that > > would help. > > > > With the equation e^5x = 1/e^2x + 7 [e is the mathematical e = 2.718] > > > > If I use Solve[e^5x == 1/e^2x + 7] I do not get an appropriately traditional > > answer. The RHS of the equation is the mathematical "e" to the power of 2x > > + 7. > > > > Using the equation e^5x = 1/e^2x + 7 It appears to me Mathematica is > > skipping the step > > > > e^5x = e^-(2x + 7 ) > > > > which would result in > > > > 5x = -(2x + 7) > > 5x = -2x - 7 > > 7x = -7 > > x = -1 > > > > Because with Mathematica I kept getting the answer: > > > > Out[1]= \!\({{x \[Rule] Log[Root[\(-1\) + 7\ #1\^5 + #1\^7 &, 1]]}, {x > > \[Rule] Log[Root[\(-1\) + 7\ #1\^5 + #1\^7 &, 2]]}, {x \[Rule] > > Log[Root[\(-1\) + 7\ #1\^5 + #1\^7 &, 3]]}, {x \[Rule] Log[Root[\(-1\) + 7\ > > #1\^5 + #1\^7 &, 4]]}, {x \[Rule] Log[Root[\(-1\) + 7\ #1\^5 + #1\^7 &, > > 5]]}, {x \[Rule] Log[Root[\(-1\) + 7\ #1\^5 + #1\^7 &, 6]]}, {x \[Rule] > > Log[Root[\(-1\) + 7\ #1\^5 + #1\^7 &, 7]]}}\) > > > > Before trying all of the above, I loaded: > > << Graphics`Graphics` > > << Algebra`AlgebraicInequalities` > > << Algebra`InequalitySolve` > > << Algebra`RootIsolation` > > > > Try restricting the problem to the Real numbers first by loading "Miscellaneous`RealOnly`" in order to avoid the complex periodicity of your original problem. Hope this helps. Jonathan Rockmann mtheory at msn.com

**References**:**Re: Exponential Equations***From:*Erich Mueller <emuelle1@uiuc.edu>