[Date Index] [Thread Index] [Author Index]

Re: Need help with solving problem

• To: mathgroup at smc.vnet.net
• Subject: [mg5398] Re: [mg5339] Need help with solving problem
• From: Perapong Tekasakul <c530789 at showme.missouri.edu>
• Date: Thu, 5 Dec 1996 14:50:32 -0500
• Sender: owner-wri-mathgroup at wolfram.com

I think you have it all right. Nothing wrong.
Log(2/3)/Log(2) is another answer.

Pong

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