Re: Exponential Equations
- To: mathgroup at smc.vnet.net
- Subject: [mg29634] Re: [mg29611] Exponential Equations
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Fri, 29 Jun 2001 01:36:07 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Using the correct syntax is the answer here:
In[20]:=
Solve[E^(5*x) == 1/E^(2*x + 7), x]//FullSimplify
Solve::ifun: Inverse functions are being used by Solve, so some solutions
may \
not be found.
Out[20]=
6 I Pi 2 I Pi
{{x -> -1}, {x -> -1 - ------}, {x -> -1 + ------},
7 7
4 I Pi 4 I Pi
{x -> -1 - ------}, {x -> -1 + ------},
7 7
2 I Pi 6 I Pi
{x -> -1 - ------}, {x -> -1 + ------}}
7 7
Note that in addition to the real solution -1 you get a few complex
solutions and indeed, as the warning says, not all the solutions have been
found. In fact there are infinitely many: -1+ 2Pi *I*n/7 is a solution for
any integer n. Of course -1 is the only real one.
--
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
http://sigma.tuins.ac.jp/~andrzej/
on 01.6.28 6:28 PM, Shippee, Steve at SHIS235 at LNI.WA.GOV 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`
>
>