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Re: [mg5339] Need help with solving problem
Posted:
Dec 1, 1996 11:21 PM


> From mathgroupadm@smc.vnet.net Wed Nov 27 16:34:00 1996 > From: ingimar.volundarson@mailbox.swipnet.se (Ingimar V\vlundarson) > To: mathgroup@smc.vnet.net > Subject: [mg5339] Need help with solving problem > ContentLength: 399 > XLines: 24 > > 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 =>
Yes ? really that are new kind of calculation rules. I think
(4^x)== 2^(2x) ==2^x*2^x and You get y^2+y2==0
finaly one gets the correct solution x> I (PiI*Log[2])/Log[2]
Hope that helps Jens



