Re: Need help with solving problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg5364] Re: [mg5339] Need help with solving problem*From*: Lou Talman <me at talmanl.mscd.edu>*Date*: Thu, 5 Dec 1996 14:50:12 -0500*Sender*: owner-wri-mathgroup at wolfram.com

Ingimar V\vlundarson wrote: > 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? Yes. Your first step is incorrect because 4^x is not the same as 2*(2^x). In fact, 4^x = (2^2)^x = 2^(2x) = (2^x)^2, so your equation is quadratic in 2^x: (2^x)^2 + (2^x) - 2 = 0. The second solution of the quadratic requires 2^x = -2, which is not possible for real x. --Lou Talman