anonymous
  • anonymous
e^log3 - e^log 2 =
Mathematics
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schrodinger
  • schrodinger
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myininaya
  • myininaya
e^log3-e^log2=3-2=1
myininaya
  • myininaya
thats assuming thats log base e
anonymous
  • anonymous
oh i see is it the same as doing this e^log3 - e^log 2 = e^log1 =e^0 =1

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myininaya
  • myininaya
e^x and lnx are inverse functions so e^ln3=3 and e^ln2=2
anonymous
  • anonymous
oh so they cancel out
myininaya
  • myininaya
yea! are we to assume that the logs are base e?
anonymous
  • anonymous
yes it says it in the text book also one other question \[\int\limits_{\log6}^{\log4} e ^{-x}\]
myininaya
  • myininaya
the antiderivative of e^(-x) is -e^(-x)+C
anonymous
  • anonymous
what ive done is this [\[[-e^{-x}\]
anonymous
  • anonymous
so you get -e^-log6+e^-log4
anonymous
  • anonymous
then using what you said before the answer should be 10
myininaya
  • myininaya
no upper limit is log4 right? lower limit is log6 right? it looks like you plugged in the lower limit first
myininaya
  • myininaya
and also if we simplify what you have we have -6^(-1)+4^(-1)=......
myininaya
  • myininaya
=-1/6+1/4
anonymous
  • anonymous
\[\int\limits_{\log4}^{\log6}e ^{-x}\] it should be that lol sorry i did it the wrong way round.
myininaya
  • myininaya
ok then what you have is right but you still simplified wrong
myininaya
  • myininaya
do you remember rlogx=logx^r
anonymous
  • anonymous
yeah my answer is wrong according to the back of the book the answer should be 1/12
anonymous
  • anonymous
yes i rember that
myininaya
  • myininaya
so we have -e^[-log6]=-e^log6^(-1)=-6^(-1)=-1/6 and we have e^[-log4]=e^log4^(-1)=4^(-1)=1/4
myininaya
  • myininaya
and, you still there?
myininaya
  • myininaya
do you got it from here? i'm fixing to go to bed
anonymous
  • anonymous
ok yeah thanks i get it lol thanks :) good night

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