I'm having a bit of trouble with this question: e2x + 3ex - 10 = 0 Thanks
passion89 Member Joined Apr 10, 2006 Messages 905 Location Outside your house Gender Female HSC 2006 Jul 3, 2006 #1 I'm having a bit of trouble with this question: e2x + 3ex - 10 = 0 Thanks
Boxxxhead Local hairy ethnic man Joined May 6, 2005 Messages 512 Location Fairfield Gender Male HSC 2006 Jul 3, 2006 #2 e2x + 3ex - 10 = 0 let m = e^x m^2 + 3m - 10 = 0.. factorises to (m + 5)(m - 2).. then you sub e^x back in (e^x + 5)(e^x - 2) = 0... e^x = -5 and e^x = 2... e^x must be > 0 so e^x = 2... convert that to log form... therefore x = ln2
e2x + 3ex - 10 = 0 let m = e^x m^2 + 3m - 10 = 0.. factorises to (m + 5)(m - 2).. then you sub e^x back in (e^x + 5)(e^x - 2) = 0... e^x = -5 and e^x = 2... e^x must be > 0 so e^x = 2... convert that to log form... therefore x = ln2
passion89 Member Joined Apr 10, 2006 Messages 905 Location Outside your house Gender Female HSC 2006 Jul 4, 2006 #3 Wow it looks so easy now...thanks heaps!
passion89 Member Joined Apr 10, 2006 Messages 905 Location Outside your house Gender Female HSC 2006 Jul 4, 2006 #4 I have one more question which I don't understand: cosө= 1/(√2) for 0<ө<2pi