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Nick88888888

  • one year ago

Write as one logarithm with base b: 3 logb x + 2 logb y

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  1. Nick88888888
    • one year ago
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    |dw:1437291202065:dw|

  2. UsukiDoll
    • one year ago
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    we need our three log rules \[\log_b(mn) =\log_bm+\log_bn\] \[\log_b(\frac{m}{n}) =\log_bm-\log_bn\] \[\log_b(m^n) =n \cdot \log_bm\]

  3. UsukiDoll
    • one year ago
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    so for this problem we don't need the subtraction one.

  4. UsukiDoll
    • one year ago
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    so anything to do with addition as seen in your problem will become multiplication as the end result... but first we need to use the third rule because you got exponents

  5. UsukiDoll
    • one year ago
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    |dw:1437293044978:dw| so use the third rule first on this

  6. UsukiDoll
    • one year ago
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    use \[\log_b(m^n) =n \cdot \log_bm \] which is the log rule for exponents first on your drawing

  7. Nick88888888
    • one year ago
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    would using Log x + Log y - Log z = Solution: Log (xy)/z work? Yea i know it doesnt have z but would this method work?

  8. UsukiDoll
    • one year ago
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    ummm yeah but where is the base though? like ok... let's have the base be b \[Log_b X + Log_b Y - Log_bZ \] then it's a combination of the first and second log rule. Your final answer should look like \[Log_b(\frac{xy}{b})\] because we have a + sign which means multiplication according to the log rules and we hae a - sign which means division according to the log rules as well.

  9. UsukiDoll
    • one year ago
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    OMG apologizes typing too fast \[Log_b(\frac{XY}{Z})\]

  10. Nick88888888
    • one year ago
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    Logb(3(2)/b)?

  11. UsukiDoll
    • one year ago
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    but you just have addition for this though so the second rule isn't necessary |dw:1437293682407:dw|

  12. UsukiDoll
    • one year ago
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    on top of that \[\log_b(m^n) =n \cdot \log_bm \] <-- this exponent rule comes first \[\log_b(mn) =\log_bm+\log_bn \] <-- this addition rule comes second

  13. UsukiDoll
    • one year ago
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    |dw:1437293771745:dw|

  14. UsukiDoll
    • one year ago
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    \[\log_b(m^n) =n \cdot \log_bm \] use the left side of this rule m = x n = 3 for the first m = y n = 2 for the second

  15. UsukiDoll
    • one year ago
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    I'll do one and you can do the other one using exponent rule for logs |dw:1437294023458:dw|

  16. Nick88888888
    • one year ago
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    |dw:1437294470191:dw|

  17. UsukiDoll
    • one year ago
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    alright.... now we just need the first exponent rule... so since we have addition our final result should be multiplication |dw:1437294541977:dw| fill in the blank

  18. Nick88888888
    • one year ago
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    |dw:1437294629036:dw|

  19. UsukiDoll
    • one year ago
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    whoa we do not combine exponents... but the xy part is correct.

  20. UsukiDoll
    • one year ago
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    |dw:1437294775553:dw|

  21. UsukiDoll
    • one year ago
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    the exponents couldn't be combined anyway due to different variables

  22. Nick88888888
    • one year ago
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    so its just logb(x^3y^2)?

  23. UsukiDoll
    • one year ago
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    yes |dw:1437294894273:dw|

  24. Nick88888888
    • one year ago
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    thank you

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