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swiftskier96

Help!?!?!?!? For questions 1 and 2, write the expression as a single logarithm. (Problem written below) logb (q^2 + t^8) logb(q^2t^8) logb(qt^(2+8)) In this answer, the 2+8 is the entire exponent. (2 + 8) logb (q + t)

  • one year ago
  • one year ago

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  1. swiftskier96
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    \[ 2\log_{b}q + 8\log_{b}t\]

    • one year ago
  2. swiftskier96
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    logb (q^2 + t^8) logb(q^2t^8) logb(qt^(2+8)) In this answer, the 2+8 is the entire exponent. (2 + 8) logb (q + t)

    • one year ago
  3. swiftskier96
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    2. 4 log x – 6 log (x + 2) (1 point) 24 log x/x + 2 log x^4(x + 2)^6 log x (x + 2)^24 none of these

    • one year ago
  4. phi
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    I don't see what you can do with the first one. for the 2nd one, can't you bring the 4 inside the log using \[ \log(x^b)= b\log(x) \] in the opposite order same for the 6 then you can replace the subtraction of logs with division inside the log

    • one year ago
  5. phi
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    oh, I see what the first question is: make \[ 2\log_{b}q + 8\log_{b}t \] into one log it is the same problem as the other one. bring the 2 inside. bring the 8 inside then change + logs to multiplication inside the log

    • one year ago
  6. swiftskier96
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    Is #2 B?

    • one year ago
  7. phi
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    B is close, but the -6 is ruining it. you change (you really should try doing this) the expression from 4 log x – 6 log (x + 2) to log x^4 - log( (x+2)^6 ) if we were adding then B would be the answer. but the - means you divide log ( x^4/ (x+2)^6 ) or using negative exponents \[\log(x^4 (x+2)^{-6})\] so unless there is a typo, B is not it.

    • one year ago
  8. androidonyourface
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    c?

    • one year ago
  9. swiftskier96
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    So if you divide the log, it would be A? (Sorry if its wrong. Im really trying my best to understand.)

    • one year ago
  10. androidonyourface
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    @phi

    • one year ago
  11. phi
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    I don't think the answer is among your choices move the 4 and the 6 inside the log \[ 4 \log x – 6 \log (x + 2) \] becomes \[ \log( x^4) - \log( (x+2)^6 ) \] the - means you divide \[ \log \left( \frac{x^4}{ (x+2)^6} \right) \]

    • one year ago
  12. swiftskier96
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    Oh ok. So it would have to be D.

    • one year ago
  13. phi
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    yes. can you take a shot at the first problem?

    • one year ago
  14. swiftskier96
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    I can try. Hold on.

    • one year ago
  15. swiftskier96
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    Would #1 be B?

    • one year ago
  16. phi
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    \[ 2\log_{b}q + 8\log_{b}t \] bring the 2 and and 8 inside (all base b, but I'm not typing it in) \[ \log(q^2) + \log(t^8) \] adding logs changes to multiplication inside a log (in the old days people did this the other way round, because adding is easier then multiplying long numbers) \[ \log(q^2t^8) \]

    • one year ago
  17. swiftskier96
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    Hey, i got it right! Lol Thanks!!! :)

    • one year ago
  18. phi
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    yw

    • one year ago
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