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anonymous

  • one year ago

Help me Simplify 9log9(7)

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  1. Owlcoffee
    • one year ago
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    \[\log_{a}B^n=nlog_{a} B\]

  2. anonymous
    • one year ago
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    9(log(9))(7) 60.117278 if this is wrong fell free to correct me

  3. geerky42
    • one year ago
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    \(\log_ba = \dfrac{\ln a}{\ln b}\), I guess.

  4. geerky42
    • one year ago
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    @aloud That's not simplifying though.

  5. anonymous
    • one year ago
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    I i think it is 9^7

  6. geerky42
    • one year ago
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    I don't think you should move 9 to exponent in log. Doesn't seem like "simplifying" to me.

  7. geerky42
    • one year ago
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    Not sure, To me, your expression is already simplified. Unless you can do something about base.

  8. anonymous
    • one year ago
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    after reworking it i get 7

  9. Owlcoffee
    • one year ago
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    Neither is transforming into a neiperian logarithm with a denominator which complicates the domain.

  10. geerky42
    • one year ago
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    "complicates the domain" Do you even know what domain is?

  11. anonymous
    • one year ago
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    Guys please i need help with this not argueing

  12. Owlcoffee
    • one year ago
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    You can hardly simplify a logarithm, you are very far from lecturing me.

  13. geerky42
    • one year ago
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    Tell me, genius. How does simplifying logarithm change its domain?

  14. geerky42
    • one year ago
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    @ganeshie8

  15. anonymous
    • one year ago
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    Please help me......

  16. geerky42
    • one year ago
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    |dw:1434463121313:dw|

  17. Owlcoffee
    • one year ago
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    In spite of geerky, who clearly hasn't mastered basic algebra yet. We will use the property I stated above: \[\log_{a} B^n=nlog_{a} B\] So, for the problem: \[9\log_{9} (7)\] We'll just take it to the exponent: \[\log_{9}(7)^9 \] That would be a more simplified version, because later on, you'll learn that log(a)B^n is a more simplified way of expressing a logarithm because of the very definition: \[\log_{a}b^n=c <=>a^c=b^n\]

  18. geerky42
    • one year ago
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    We are simplifying a expression, not equation. So turn it into exponential form wouldn't work out well. Plus it's not \(\log_9(7)^9\), it's \(\log_9(7^9)\)

  19. geerky42
    • one year ago
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    Multiplication looks simplified to me than exponent. So as far as I concern, given expression is pretty much already simplified. Exactly what kind of form are you asked to simplify into? @maryah.g.osborn

  20. Owlcoffee
    • one year ago
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    Do as you desire...

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  21. geerky42
    • one year ago
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    http://www.wolframalpha.com/input/?i=9+log+base+9+%287%29

  22. geerky42
    • one year ago
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    /thread

  23. Owlcoffee
    • one year ago
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    hahahaha... As I said, do as you deem fit, I have no desire to argue with unmature people.

  24. geerky42
    • one year ago
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    Clearly because you don't want to make yourself look any more fool.

  25. ganeshie8
    • one year ago
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    relax friends i think the question looks terrible because of not using latex, i think it should be \[\large 9^{\log_9 7}\] @maryah.g.osborn please confirm if the expression looks as above or attach a screenshot if possible

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