Re: I don't see what I am doing wrong

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: t= ln 1 = 0 : and you are right, I got confused. Formatting is very good. Thanx :) : : : (e^(2t))-(3e^t)+2=0 : : : let x =e^t : : : so (x^2)-(3x)+2=0 : : : (x-2)(x-1) : : : ((e^t)-2)((e^t)-1) : : : so ln^(et) = ln2 : : : and ln^(et) = 1 : : : and that is wrong, the answer is suppose to be 0 and ln2 : : : What did I do wrong? : : So close! : : Let me start at: : : : : (e^t-2)(e^t-1)=0 : : Note You left out the =0. When you work problems always make sure that : : everything that is written down is a complete true statement. It will : : help you come back and study the problem letter. : : I will break the equation in to two parts but I will only do one part: : : (e^t-2)=0 : : e^t=2 : : ln{e^t}=ln{2} : : Note ln{e^t}=t therefore: : : t=ln{2} : : Now you do: : : (e^t-1)=0

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