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swiftskier96
Group Title
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
swiftskier96 Group Title
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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swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
\[ 2\log_{b}q + 8\log_{b}t\]
 one year ago

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
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

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
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

phi Group TitleBest ResponseYou've already chosen the best response.1
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

phi Group TitleBest ResponseYou've already chosen the best response.1
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

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
Is #2 B?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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

androidonyourface Group TitleBest ResponseYou've already chosen the best response.0
c?
 one year ago

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
So if you divide the log, it would be A? (Sorry if its wrong. Im really trying my best to understand.)
 one year ago

androidonyourface Group TitleBest ResponseYou've already chosen the best response.0
@phi
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
Oh ok. So it would have to be D.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes. can you take a shot at the first problem?
 one year ago

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
I can try. Hold on.
 one year ago

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
Would #1 be B?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
\[ 2\log_{b}q + 8\log_{b}t \] 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

swiftskier96 Group TitleBest ResponseYou've already chosen the best response.1
Hey, i got it right! Lol Thanks!!! :)
 one year ago
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