A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

what is the natural log of e^-x? Can anyone help?

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -x

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ln(e^x)=x\] because the log is the inverse of the exponential.

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    likewise \[e^{ln(x)}=x\]

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok let me write it.

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks.

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i think the simplest thing to do is remember that \[e^{-x}=\frac{1}{e^x}\]

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    actually first lets do a simple trick. multiply by \[e^x\] to get \[e^x(e^x-11e^{-x})=0\] \[e^{2x}-11e^{-x+x})=0\] \[e^{2x}-11e^0=0\] \[e^{2x}-11=0\]

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hope that was not too many steps i just wanted to make sure it was clear. now we solve for x: \[e^{2x}=11\] \[2x=ln(11)\] \[x=\frac{ln(11)}{2}\]

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you thank you thank you in advance for taking the time!

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thankyou! what about solving e^x + e^-x - 12 for x? I jusgt really don't get this stuff. Your help is wonderful!

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is that \[e^x+e^{-x}=12\]?

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's e^x + e^-x -12 = 0

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well this one is a pain. first of all multiply by \[e^x\]

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    get \[ e^{2x}+1-12e^x=0\] or \[e^{2x}-12e^x+1=0\]

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hope it is clear that it is a quadratic.

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe i messed up. oh no. ok here goes.

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[e^{2x}-12e^x+1=0\] quadratic equation in \[e^x\] so solve \[z^2-12z+1=0\] using the quadratic formula and then replace z by \[e^x\]

  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i just want to know how yhou know all this stuff?

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[z^2-12z+1=0\] \[z^2-12z=-1\] \[(z-6)^2=35\] \[z-6=\pm\sqrt{35}\] \[z=6\pm \sqrt{35}\]

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i didn't use the quadratic formula i completed the square. same thing but it keeps me from having to simplify the radical

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you will get the same answer with formula, it will just take longer.

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    omg thanks again. i am writing frantically to keep up, but i am not even close to your abilities!

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[e^{x}=6+\sqrt{35}\] \[x=ln(6+\sqrt{35})\] \ \[x=ln(6-\sqrt{35})\]

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well it is written here. if you have any questions about any steps ask

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Again, thanks and maybe I will talk to you again soon if you have another 3 hours for another problem! : )

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good luck with calc!

  33. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.