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

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Posted by Brad Paul on October 11, 2002 at 07:39:41:

In Reply to: I don't see what I am doing wrong posted by Lisa on October 11, 2002 at 02:35:51:

: (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

Follow Ups:

Re: I don't see what I am doing wrong Lisa 14:45:15 10/11/02 (2)
- question..... rk 14:58:53 10/11/02 (1)
  - Re: question..... Brad Paul 15:39:42 10/11/02 (0)

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Comments:
: : (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: : : (et-2)(et-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: : (et-2)=0 : et=2 : ln{et}=ln{2} : Note ln{et}=t therefore: : t=ln{2} : Now you do: : (et-1)=0

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