2 2323 Member Joined Apr 5, 2012 Messages 40 Gender Female HSC 2013 Aug 12, 2012 #1 Does anyone know how to solve 2^2x -6(2^x) + 8 = 0
F falsenthusiasm Member Joined Sep 15, 2011 Messages 99 Gender Undisclosed HSC 2013 Aug 12, 2012 #2 ((2^x)^2) - 6(2^x) + 8 = 0 Let A = 2^x Think you can do the rest?
Sendoh08 Member Joined Feb 15, 2011 Messages 106 Gender Male HSC 2012 Aug 30, 2012 #3 ^ Yup whay he said: ((2^x)^2) - 6(2^x) + 8 = 0 A^2 - 6A + 8 = 0 (A - 4)(A - 2) = 0 (Factorising above line) A = 4 or A = 2 2^x = 4 or 2^x = 2 therefore x=2 or x=1 And thats the working and solution =)
^ Yup whay he said: ((2^x)^2) - 6(2^x) + 8 = 0 A^2 - 6A + 8 = 0 (A - 4)(A - 2) = 0 (Factorising above line) A = 4 or A = 2 2^x = 4 or 2^x = 2 therefore x=2 or x=1 And thats the working and solution =)