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anonymous

  • 4 years ago

Expand the Expression x>0 , y>0 , z>0 log (x^2/yz^4)

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  1. amistre64
    • 4 years ago
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    log a/b = loga - logb log ab = loga + logb

  2. amistre64
    • 4 years ago
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    log n^r = r logn

  3. anonymous
    • 4 years ago
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    him can you elaborate more please

  4. amistre64
    • 4 years ago
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    not really, those are the basic rules for log operations. I can complicate them, but I doubt I can make them any easier ...

  5. anonymous
    • 4 years ago
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    can you giveme an example and work it out

  6. amistre64
    • 4 years ago
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    ok, but its just gonna be the same thing but with numbers in there

  7. amistre64
    • 4 years ago
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    log (1/2) = log(1)-log(2) log(4*3) = log(4)+log(3) log(6^3) = 3 log(6)

  8. amistre64
    • 4 years ago
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    youll need all these operations to expand the one youve got

  9. anonymous
    • 4 years ago
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    so the final answer would be 3log(6)?

  10. anonymous
    • 4 years ago
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    loga (x^2/yz^4)--what do i do with a

  11. amistre64
    • 4 years ago
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    "a" stays the same unless your asked to change of base it

  12. amistre64
    • 4 years ago
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    log (x^2/yz^4) do you see the division sign? the fraction bar? split this into its subtraction parts

  13. anonymous
    • 4 years ago
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    yesI understand it now! thanks

  14. amistre64
    • 4 years ago
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    log (x^2/yz^4) = log (x^2) - log (yz^4) where you see the multiplication, split it into its addition parts log (x^2/yz^4) = log (x^2) - log (y)+ log(z^4) and where you see exponents, turn it into its like parts

  15. amistre64
    • 4 years ago
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    log (x^2/yz^4) = 2log (x) - log (y)+ 4log(z)

  16. amistre64
    • 4 years ago
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    the base "a" doesnt change thruout it unless directed otherswise

  17. amistre64
    • 4 years ago
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    i might have a typo in there :)

  18. amistre64
    • 4 years ago
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    log (x^2/yz^4) = log (x^2) - (log (y)+ log(z^4)) is better since the hole of it is subtracted log (x^2/yz^4) = log (x^2) - log (y) - log(z^4) and then its the same from there

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