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MathGroup Archive 1996

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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!
> 
> 
> 


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