Re: Exponential Equations
- To: mathgroup at smc.vnet.net
- Subject: [mg29645] Re: Exponential Equations
- From: "Bill Bertram" <wkb at ansto.gov.au>
- Date: Fri, 29 Jun 2001 01:36:22 -0400 (EDT)
- Organization: Australian Nuclear Science and Technology Organisation
- References: <9hetrp$6ku$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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. That's because in your equation you don't have "e" to the power of 2x + 7. What you have been trying to solve is e^(5x) == 1/(e^(2x) + 7) What you really want is e^(5x) == 1/(e^(2x+7)) and for this Mathematica does indeed yield a solution x -> -1. Cheers, Bill