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