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 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` >