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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`
>
>


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