A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

StudyGurl14

  • one year ago

HELP PLEASE Solving for x.

  • This Question is Closed
  1. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1440620935256:dw| Obviously, I did something wrong.

  2. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @perl

  3. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1440621059469:dw|

  4. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I see. I think.

  5. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    multiply both sides by 2^x, to get rid of 2^x in the denominator

  6. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1440621096683:dw| Like that?

  7. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1440621179162:dw|

  8. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hint: you will wind up with a quadratic in terms of 2^x

  9. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1440621199302:dw|

  10. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I forgot the 1.

  11. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so...? |dw:1440621240821:dw|

  12. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ???

  13. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now do I use ln?

  14. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1440621466522:dw|

  15. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, so 2^2x

  16. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1440621493371:dw|

  17. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1440621524603:dw| ???

  18. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Now you have a quadratic equation.

  19. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How?^

  20. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    FYI, what you did wrong is that when you start with \[ 2^x + 2^{-x}= 5 \] we take the ln of both sides \[ \ln \left( 2^x + 2^{-x}\right) = \ln 5\] and that is *not* \[ \ln \left( 2^x\right) + \ln \left( 2^{-x}\right) = \ln 5\]

  21. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, so how do I distrubite the ln @phi ?

  22. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    if we let u = 2^x. then we have u^2 - 5*u +1 = 0

  23. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you can't distribute the ln

  24. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I see @perl

  25. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so.... 5/2 +- sqrt{21}/2 ?

  26. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, and now you can do 2^x = mess and take the log of both sides

  27. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what's "mess" ?

  28. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    5/2 +- sqrt{21}/2

  29. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay, so |dw:1440621917929:dw| like that?

  30. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    we want to solve (2^x)^2 - 5*(2^x) +1 = 0 if we let u = 2^x. then we have u^2 - 5*u +1 = 0 the solutions are u = 5/2 + sqrt{21}/2 , 5/2 - sqrt{21}/2 Now back substitute since we really wanted to solve for x 2^x = 5/2 + sqrt{21}/2 2^x = 5/2 - sqrt{21}/2 now take ln of both sides ln (2^x)= ln( 5/2 + sqrt{21}/2)

  31. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got approx 2.26?

  32. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes

  33. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay, thank you.

  34. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and the other solution?

  35. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do I do it again for the negative one?

  36. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    right

  37. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so approx -2.26

  38. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    indeed :)

  39. StudyGurl14
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you!

  40. perl
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and as phi pointed out, in general ln( x + y) does not equal ln(x) + ln (y) same with other many other functions sin ( x + y) does not equal sin(x) + sin(y),

  41. 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.