Re: Need help with solving problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg5363] Re: [mg5339] Need help with solving problem*From*: Robert Pratt <rpratt at math.unc.edu>*Date*: Thu, 5 Dec 1996 14:50:11 -0500*Sender*: owner-wri-mathgroup at wolfram.com

Yes, you are doing something wrong. 4^x does not equal 2*2^x, as you can check by letting x=2. However, 4^x=(2*2)^x=(2^x)(2^x)=(2^x)^2. So let y=2^x and solve y^2+y-2=0. You get y=-2 or y=1, hence 2^x=-2 or 2^x=1. The first equation yields no solution, but the second gives x=0. Rob Pratt Department of Mathematics The University of North Carolina at Chapel Hill CB# 3250, 331 Phillips Hall Chapel Hill, NC 27599-3250 rpratt at math.unc.edu On Wed, 27 Nov 1996, Ingimar Vvlundarson wrote: > Can anyone help me with this problem? > > (4^x)+(2^x)-2=0 > > Everyone can see that x=0 but: > > (4^x)+(2^x)-2=0 => > 2*(2^x)+(2^x)-2=0 => > 2y+y-2=0 => ;(let y=2^x) > 3y=2 => > y=2/3 => > 2^x=2/3 => ;(let 2^x=y) > lg(2^x)=lg(2/3) => > x*lg(2)=lg(2/3) => > x=lg(2/3)/lg(2) Which is not true!!! > > Am I doing something wrong? > > Please E-mail an answer to: > ingimar.volundarson at mailbox.swipnet.se > > Thanx in advance! > > >