Re: Exponential Equations
- To: mathgroup at smc.vnet.net
- Subject: [mg29626] Re: Exponential Equations
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Fri, 29 Jun 2001 01:35:59 -0400 (EDT)
- References: <9hetrp$6ku$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Steve,
We have to use parentheses to specify grouping:
Solve[E^(5x)==1/E^(2x+7),x]
Solve::ifun: Inverse functions are being used by Solve, so some
solutions may \
not be found.
{{x -> -1}, {x -> Log[-((-1)^(1/7)/E)]},
{x -> Log[(-1)^(2/7)/E]}, {x -> Log[-((-1)^(3/7)/E)]},
{x -> Log[(-1)^(4/7)/E]}, {x -> Log[-((-1)^(5/7)/E)]},
{x -> Log[(-1)^(6/7)/E]}}
FullSimplify[%]
{{x -> -1}, {x -> -1 - (6*I*Pi)/7},
{x -> -1 + (2*I*Pi)/7}, {x -> -1 - (4*I*Pi)/7},
{x -> -1 + (4*I*Pi)/7}, {x -> -1 - (2*I*Pi)/7},
{x -> -1 + (6*I*Pi)/7}}
The extra solutions are because Mathematica looks for complex solutions. It
finds some, but not all, of them.
We have
E^(5x) == E^-(2x + 7 )
5x ==(-2x+7) + 2n Pi I (*n an integer*)
7x == 7+ 2n Pi I
x == 1+ 2n Pi I/7
However we do have
<<Algebra`InequalitySolve`
InequalitySolve[E^(5x)==1/E^(2x+7),x]
InequalitySolve::npi: A nonpolynomial equation or inequality
encountered. The solution set may be incorrect.
x==-1
<<Miscellaneous`RealOnly`
Solve[E^(5x)==1/E^(2x+7),x]
Solve::ifun: Inverse functions are being used by Solve, so
some solutions may not be found.
Nonreal::warning: Nonreal number encountered.
{{x->-1},{x->Nonreal}}
Instead of the standard package RealOnly, you might like to use
http://www.mathsource.com/Content/Enhancements/MathFunctions/0211-396
which gives a version that can be turned off after use.
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Shippee, Steve" <SHIS235 at LNI.WA.GOV> wrote in message
news:9hetrp$6ku$1 at smc.vnet.net...
> 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`
>