       Re: Exponential Equations

• To: mathgroup at smc.vnet.net
• Subject: [mg29624] Re: Exponential Equations
• From: "Orestis Vantzos" <atelesforos at hotmail.com>
• Date: Fri, 29 Jun 2001 01:35:57 -0400 (EDT)
• Organization: National Technical University of Athens, Greece
• References: <9hetrp\$6ku\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Priority of operators, my friend (highschool stuff):
e^5x==1/e^2x+7 is NOT e^5x==e^-(2x+7)
1/e^2x+7 == (1/e^2x)+7 != 1/e^(2x+7) which is what you want...
Orestis
PS.Mathematica unfortunately does not have the capability to "guess" your
problem....yet;-)

"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
> 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= \!\({{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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