Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

BelleFlower

  • 3 years ago

Logarithms! Wait for image please!

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

    Part c!

    1 Attachment
  2. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hint : write \(2=\log_2 2^2\)

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

    This is what I have done so far: log_2 (x - 1)^2 = log_2 4 + log_2 (x+2) (x-1)^2 = 4 + x + 2 x^2 - 3x - 7 = 0

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

    Is that correct?

  5. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no....

  6. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (x-1)^2 = 4 ( x + 2 ) u got how ?

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

    Oh wait is it: (x - 1)^2 = 4x + 8 or am I still wrong?

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

    @hartnn

  9. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now correct.

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

    So if I solve that I should be able to get the answer right?

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

    Okay thanks! :)

  12. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes, welcome ^_^

  13. 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