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MathGroup Archive 1996

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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




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