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Anikate

  • one year ago

use property of logariths to completely expand: (posting equation)

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  1. Anikate
    • one year ago
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    \[\ln ((x^18 \]

  2. Anikate
    • one year ago
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    ignor that

  3. Anikate
    • one year ago
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    |dw:1434341420293:dw|

  4. Anikate
    • one year ago
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    and yes

  5. mathstudent55
    • one year ago
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    Here are the log rules you need: |dw:1434341768384:dw|

  6. Anikate
    • one year ago
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    yea its cool, I can sorta see what hes doing, equation maker would help though.

  7. Anikate
    • one year ago
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    but the equation is in LN...... those equations are in log

  8. UsukiDoll
    • one year ago
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    \[\ln \frac{x^{18}(y+3)^7}{z}\]

  9. UsukiDoll
    • one year ago
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    ln and log mean the same thing

  10. mathstudent55
    • one year ago
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    Use the first 2 rules to take care of the fraction. Then use the third rule to deal with the exponents.

  11. Anikate
    • one year ago
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    oh ok

  12. mathstudent55
    • one year ago
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    When you see the log of a product, separate it into the sum of logs. When you see the log of a division, separate it into a subtraction of logs.

  13. UsukiDoll
    • one year ago
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    we could split this problem into parts and then combine them all together...

  14. mathstudent55
    • one year ago
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    |dw:1434342091696:dw|

  15. Anikate
    • one year ago
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    how am i supposed to put all those equations together to get the answer?

  16. UsukiDoll
    • one year ago
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    take it one step at a time using the log rules.

  17. mathstudent55
    • one year ago
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    |dw:1434342139208:dw|

  18. mathstudent55
    • one year ago
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    |dw:1434342185238:dw|

  19. Anikate
    • one year ago
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    thanks @HWBUSTER00 i feel like I have more questions than mathway can tell me lol thanks for ur help though :)

  20. Anikate
    • one year ago
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    @mathstudent55 you cant expand it more?

  21. UsukiDoll
    • one year ago
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    I think we still can for the first part... it's a combination of the exponential log rule and the addition rule

  22. Anikate
    • one year ago
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    yea... how though? wouldnt we have to use brackets?

  23. UsukiDoll
    • one year ago
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    |dw:1434342343219:dw|

  24. UsukiDoll
    • one year ago
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    it's a combination of the first and third log rule.. .

  25. UsukiDoll
    • one year ago
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    |dw:1434342367152:dw|

  26. UsukiDoll
    • one year ago
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    so I KNOW! We use the first log rule and then the third log rule

  27. mathstudent55
    • one year ago
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    The first step was using the log of a fraction. Now we use the log of a product on the first log.

  28. UsukiDoll
    • one year ago
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    use the first log rule he gave out...

  29. mathstudent55
    • one year ago
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    |dw:1434342473232:dw|

  30. UsukiDoll
    • one year ago
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    YES ^ and then use the exponential log rule

  31. mathstudent55
    • one year ago
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    That expanded the first log, which was the log of a product.

  32. Anikate
    • one year ago
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    can u expand lnx^18?

  33. mathstudent55
    • one year ago
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    Now you need to use the log of a power, the third rule above.

  34. mathstudent55
    • one year ago
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    |dw:1434342571914:dw|

  35. Anikate
    • one year ago
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    OH I GOT IT! I understand now! thanks!

  36. mathstudent55
    • one year ago
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    That is the final answer.

  37. Anikate
    • one year ago
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    reviewing for final, and the its all starting to come back to me

  38. mathstudent55
    • one year ago
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    |dw:1434342686828:dw|

  39. mathstudent55
    • one year ago
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    Great. Good luck.

  40. UsukiDoll
    • one year ago
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    for log rules.. anything to do with a + sign it's multiplication anything to do with a - sign it's division anything to do with an exponent goes on the left side of the log. :)

  41. Anikate
    • one year ago
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    gotcha! thanks usuki!

  42. UsukiDoll
    • one year ago
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    I'm on a roll to answering a lot of questions today ^_^ I got 606 medals xD

  43. UsukiDoll
    • one year ago
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    well that's over a course of almost 3 years

  44. Anikate
    • one year ago
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    damn...... nice!

  45. UsukiDoll
    • one year ago
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    thank you :'D

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