Exponential Equations

• To: mathgroup at smc.vnet.net
• Subject: [mg29611] Exponential Equations
• From: "Shippee, Steve" <SHIS235 at LNI.WA.GOV>
• Date: Thu, 28 Jun 2001 05:28:01 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```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`

```

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