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jennyjewell25

  • 3 years ago

Write the expression as a single natural logarithm.

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  1. jennyjewell25
    • 3 years ago
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    \[3\ln x - 2\ln c\]

  2. jennyjewell25
    • 3 years ago
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    the c is closer to the ln like lnc

  3. jennyjewell25
    • 3 years ago
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    @jim_thompson5910

  4. jim_thompson5910
    • 3 years ago
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    Hint: Use the following identities \[\Large y*\ln(x) = \ln(x^y)\] \[\Large \ln(x)-\ln(y) = \ln\left(\frac{x}{y}\right)\]

  5. jennyjewell25
    • 3 years ago
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    huh?

  6. jim_thompson5910
    • 3 years ago
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    |dw:1352158470384:dw|

  7. jim_thompson5910
    • 3 years ago
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    becomes |dw:1352158488263:dw|

  8. jennyjewell25
    • 3 years ago
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    im confused whats with the y

  9. jim_thompson5910
    • 3 years ago
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    it's just a general way of stating the rule

  10. jim_thompson5910
    • 3 years ago
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    you can replace y with any number or variable you want

  11. jennyjewell25
    • 3 years ago
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    if this is as hard as the problem from earlier im just going to guess lol

  12. jim_thompson5910
    • 3 years ago
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    examples: |dw:1352158554240:dw|

  13. jennyjewell25
    • 3 years ago
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    \[\ln x ^{3}c ^{2}\]

  14. Decart
    • 3 years ago
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    \[lnx ^{3}-lnc ^{2}\]

  15. jim_thompson5910
    • 3 years ago
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    no

  16. jim_thompson5910
    • 3 years ago
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    you're close though

  17. jennyjewell25
    • 3 years ago
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    thats what i got but im probley wrong

  18. jennyjewell25
    • 3 years ago
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    x is over c isnt it

  19. jim_thompson5910
    • 3 years ago
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    it is

  20. jim_thompson5910
    • 3 years ago
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    it should be \[\Large \ln\left(\frac{x^3}{c^2}\right)\]

  21. jennyjewell25
    • 3 years ago
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    thought so

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