       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
> 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

```

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