Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

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


  • Prev by Date: Re: Naming pieces of patterns
  • Next by Date: Thread and MapThread
  • Previous by thread: Re: Exponential Equations
  • Next by thread: fills for barchart