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MathGroup Archive 1996

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Re: Need help with solving problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg5372] Re: [mg5339] Need help with solving problem
  • From: Francisco Edmundo de Andrade <edmundo at lia.ufc.br>
  • Date: Thu, 5 Dec 1996 14:50:15 -0500
  • Sender: owner-wri-mathgroup at wolfram.com


On Wed, 27 Nov 1996, Ingimar V=F6lundarson wrote:

> Can anyone help me with this problem?
>=20
> (4^x)+(2^x)-2=3D0
>=20
> Everyone can see that x=3D0 but:
>=20
> (4^x)+(2^x)-2=3D0 =3D>
> 2*(2^x)+(2^x)-2=3D0 =3D>
> 2y+y-2=3D0 =3D>      ;(let y=3D2^x)
> 3y=3D2 =3D>
> y=3D2/3 =3D>
> 2^x=3D2/3 =3D>       ;(let 2^x=3Dy)
> lg(2^x)=3Dlg(2/3) =3D>=20
> x*lg(2)=3Dlg(2/3) =3D>
> x=3Dlg(2/3)/lg(2) Which is not true!!!
>=20
> Am I doing something wrong?
>=20
> Please E-mail an answer to:
> ingimar.volundarson at mailbox.swipnet.se
>=20
> Thanx in advance!
>=20
>=20
>=20

Yes, there is an error in the first equivalence:
(4^x)+(2^x)-2=3D0   <>  2*(2^x)+(2^x)-2=3D0
because
(4^x)+(2^x)-2=3D0   =3D   (2^x)*(2^x)+(2^x)-2=3D0

So, the sequence must be continued:
y^2+y-2=3D0        ; 2^x->y
y=3D-2 or y=3D1
2^x=3D-2 or 2^x=3D1
The first equation is invalid, so the solution is x=3D0.

(Edmundo's contribuition 1996)

With Kindest regards

Edmundo
-----------------------------
     edmundo at lia.ufc.br
Universidade Federal do Ceara
 Fortaleza - Ceara - BRAZIL
-----------------------------



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