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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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