Re: Exponential Equations
- To: mathgroup at smc.vnet.net
- Subject: [mg29640] Re: [mg29611] Exponential Equations
- From: Tomas Garza <tgarza01 at prodigy.net.mx>
- Date: Fri, 29 Jun 2001 01:36:14 -0400 (EDT)
- References: <200106280928.FAA06715@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Why not try NSolve? (BTW, I suggest you use the function Exp instead of E:
the latter is the mathematical constant e, while you need the exponential
function, right?).
In[1]:=
NSolve[Exp[5 x] == 1/Exp[2 x] + 7, x]
Solve::"ifun": "Inverse functions are being used by \!\(Solve\), so some \
solutions may not be found."
Out[1]=
{{x -> -0.972956895873466 - 1.5713472864108575*I},
{x -> -0.972956895873466 + 1.5713472864108575*I},
{x -> 0.3783560491163147 - 1.248122665490004*I},
{x -> 0.3783560491163147 + 1.248122665490004*I},
{x -> 0.39380756235429715 - 2.525262904008179*I},
{x -> 0.39380756235429715 + 2.525262904008179*I},
{x -> 0.4015865688057083}}
You get a number of complex roots, but the last one is real. You may test it
to check it is Ok:
In[2]:=
(Exp[5 x] == 1/Exp[2x] + 7) /. {x -> 0.4015865688057083`}
Out[2]=
True
You may also look at the plot to check that this seems to be the only real
root:
In[3]:=
Plot[Exp[5 x] - 1/Exp[2x] - 7, {x, -1, 1}];
Tomas Garza
Mexico City
----- Original Message -----
From: "Shippee, Steve" <SHIS235 at LNI.WA.GOV>
To: mathgroup at smc.vnet.net
Subject: [mg29640] [mg29611] Exponential Equations
> 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`
>
- References:
- Exponential Equations
- From: "Shippee, Steve" <SHIS235@LNI.WA.GOV>
- Exponential Equations