Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

SugarRainbow

  • 3 years ago

how to simplify square root of (9*e^4)*2*(e^-3)?

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

    multiply (9e^4) * 2 first. what do you get?

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

    sorry i did that wrong its like this|dw:1352271755951:dw|

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

    well that's different... take teh square root of \(\sqrt{9e^4}\) i assume you know how to?

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

    you get 3e^2?

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

    yes now multiply that to 2

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

    6e^2 or 6e^4?

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

    6e^2

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

    now multiply that to e^-3

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

    6e^-6?

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

    no

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

    remember rules of exponents.. \[\huge x^a \times x^b \implies x^{a + b}\]

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

    oh so 6e^2+-3 so 6e^-1?

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

    if you have negative exponent that just becomes \[\huge x^a \times x^{-b} \implies x^{a + (-b)} \implies x^{a - b}\]

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

    yes 6e^-1

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

    now...recall your rules of exponents again.. \[\huge x^{-m} \implies \frac 1 {x^m}\]

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

    so 6e?

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

    no. look at my comment very carefully

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

    i didn't say \[\huge x^{-m} \implies x^m\]

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

    so 6(1/e^1)?

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

    yes. simplify it more

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

    can i multiply 6 by that? so like 6/e^1?

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

    yes. simplify it more

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

    IDK how unless it's e^-1/6

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

    no

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

    6/e^1 was already right.. you just have tp simplify it more into 6/e...e^1 = e remember?

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

    so 6/e?

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

    yes

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

    is that it?

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

    yes

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

    okay thank you goodnight :)

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

    unless you think you can simplify it further

  32. 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