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anonymous

  • one year ago

please help, question below

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  1. anonymous
    • one year ago
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  2. mathstudent55
    • one year ago
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    You need to know these rules of logs: \(\Large \log_b \dfrac{x}{y} = \log_bx - \log_b y\) \(\Large \log_b x^n = n \log_b x\) Use the first rule first. What do you get?

  3. anonymous
    • one year ago
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    so that means that in the second part of the equation, it always has to be subtracting... okay. that takes 2 answers off my list.

  4. anonymous
    • one year ago
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    @mathstudent55

  5. anonymous
    • one year ago
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    @mathstudent55

  6. anonymous
    • one year ago
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    would it be \[\log_{4}8-2\log_{4}x \]

  7. anonymous
    • one year ago
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    someone please help me

  8. anonymous
    • one year ago
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    @Luigi0210

  9. radar
    • one year ago
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    You're doing just fine.

  10. anonymous
    • one year ago
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    did i get the right answer??:D

  11. anonymous
    • one year ago
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    @radar

  12. Luigi0210
    • one year ago
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    Yea you're right :_

  13. radar
    • one year ago
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    No, that would be dividing 8 squared by c

  14. radar
    • one year ago
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    I mean *x lol

  15. radar
    • one year ago
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    Look at the earlier post by mathstudent55, you are supposed to be subtracting the divisor.

  16. radar
    • one year ago
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    Did you choose D "did i get the right answer??:D"

  17. anonymous
    • one year ago
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    no, thats a super smiley face, haha. i choose C and it was correct

  18. radar
    • one year ago
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    O..K that is correct!

  19. mathstudent55
    • one year ago
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    Applying the first rule we get: \( \Large \log_4 \dfrac{8}{x^2} = \log_4 8 - \log_4 x^2\) Applying the second rule we get: \( \Large \log_4 \dfrac{8}{x^2} = \log_4 8 - 2 \log_4 x\)

  20. mathstudent55
    • one year ago
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    You are correct. Good job!

  21. radar
    • one year ago
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    Sorry about that I guess I am not "hip" to internet speak lol.

  22. radar
    • one year ago
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    I thought you were on track that is why I said you are doing fine, then you posted that and it threw me off.

  23. radar
    • one year ago
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    @mathstudent55 Just a problem in "internet slang"

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spraguer (Moderator)
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