anonymous
  • anonymous
what is the natural log of e^-x? Can anyone help?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
-x
anonymous
  • anonymous
\[ln(e^x)=x\] because the log is the inverse of the exponential.
anonymous
  • anonymous
likewise \[e^{ln(x)}=x\]

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anonymous
  • anonymous
how do i solve this one then? Do you mind helping? it's e^x - 11e^-x. I need to set it to 0 and solve for x.
anonymous
  • anonymous
ok let me write it.
anonymous
  • anonymous
thanks.
anonymous
  • anonymous
ok i think the simplest thing to do is remember that \[e^{-x}=\frac{1}{e^x}\]
anonymous
  • anonymous
actually first lets do a simple trick. multiply by \[e^x\] to get \[e^x(e^x-11e^{-x})=0\] \[e^{2x}-11e^{-x+x})=0\] \[e^{2x}-11e^0=0\] \[e^{2x}-11=0\]
anonymous
  • anonymous
hope that was not too many steps i just wanted to make sure it was clear. now we solve for x: \[e^{2x}=11\] \[2x=ln(11)\] \[x=\frac{ln(11)}{2}\]
anonymous
  • anonymous
thank you thank you thank you in advance for taking the time!
anonymous
  • anonymous
if any steps are not clear let me know. btw we can multiply by \[e^x\] because it is never 0. in fact \[e^x\] is always positive.
anonymous
  • anonymous
thankyou! what about solving e^x + e^-x - 12 for x? I jusgt really don't get this stuff. Your help is wonderful!
anonymous
  • anonymous
is that \[e^x+e^{-x}=12\]?
anonymous
  • anonymous
it's e^x + e^-x -12 = 0
anonymous
  • anonymous
try multiplying again by \[e^x\] and see what happens. i will do it with pencil because if i type on the fly i might make a mistake
anonymous
  • anonymous
do it the same way again then? It's so funny that this is a question on our test and we have never done anything like it! Is that calc one? That is way more than I am ready for after only 3 weeks of calculus one.
anonymous
  • anonymous
well this one is a pain. first of all multiply by \[e^x\]
anonymous
  • anonymous
get \[ e^{2x}+1-12e^x=0\] or \[e^{2x}-12e^x+1=0\]
anonymous
  • anonymous
this one in a quadratic equation in \[e^x\] and it doesn't factor so you have to use the quadratic formula with a = 1, b = -12 and c = 1
anonymous
  • anonymous
hope it is clear that it is a quadratic.
anonymous
  • anonymous
maybe i messed up. oh no. ok here goes.
anonymous
  • anonymous
\[e^{2x}-12e^x+1=0\] quadratic equation in \[e^x\] so solve \[z^2-12z+1=0\] using the quadratic formula and then replace z by \[e^x\]
anonymous
  • anonymous
i just want to know how yhou know all this stuff?
anonymous
  • anonymous
\[z^2-12z+1=0\] \[z^2-12z=-1\] \[(z-6)^2=35\] \[z-6=\pm\sqrt{35}\] \[z=6\pm \sqrt{35}\]
anonymous
  • anonymous
oh i didn't use the quadratic formula i completed the square. same thing but it keeps me from having to simplify the radical
anonymous
  • anonymous
you will get the same answer with formula, it will just take longer.
anonymous
  • anonymous
omg thanks again. i am writing frantically to keep up, but i am not even close to your abilities!
anonymous
  • anonymous
\[e^{x}=6+\sqrt{35}\] \[x=ln(6+\sqrt{35})\] \ \[x=ln(6-\sqrt{35})\]
anonymous
  • anonymous
well it is written here. if you have any questions about any steps ask
anonymous
  • anonymous
btw if you didn't know the trick of multiplying by \[e^x\] you could have added: \[e^x+e^{-x}=12\] \[e^x+\frac{1}{e^x}=12\] \[\frac{e^{2x}+1}{e^x}=12\] \[e^{2x}+1=e^{12x}\] and you see you get the same equation again.
anonymous
  • anonymous
Again, thanks and maybe I will talk to you again soon if you have another 3 hours for another problem! : )
anonymous
  • anonymous
good luck with calc!

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