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anonymous
 one year ago
log(p+q)(pq) = 1
log(p+q)log(p^2q^2) = ?
option: a)2 b)1 c) 1 d) 0
anonymous
 one year ago
log(p+q)(pq) = 1 log(p+q)log(p^2q^2) = ? option: a)2 b)1 c) 1 d) 0

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cud anybody give me the full solution. ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0umm I don't know if this is correct: \[\log(p+q)\log(p^2q^2) =?\] \[\log(p+q)\log(pq) = 1\] so first equation is turned into: \[\log(p+q)\log((p+q)(pq)) = ?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since \[\log(p+q)(pq) = 1\] so \[\log(p+q) 1 =0\]

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0log(p^2q^2) = log(p+q)(pq) = 1 but i cant see a way to find what log(p+q) is.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella that's in my last reply

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so since log(p+q) = 1 it will be: \[\log(p+q)\log(p^2q^2) = 0\]

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0i cant see your logic there  I'm not saying you are wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean to say that since log(p^2q^2) = 1 log(p+q)log(p^2q^2) = will be log(p+q) 1 = 0 so log(p+q) = 1

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0if log ( p + q) = 1 log(p+q)log(p^q^2) = 1 * 1 = 1 not 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yes my bad I was subtracting then yes it would be 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.0@nopen it looked like you said: \[\log(a \cdot b)=\log(a) \cdot \log(b) \text{ thisn't a log property though }\]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0can the asker post the screenshot of the question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but isn't it log(a*b) = log(a) + log(b)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0that is the product rule for log

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0I think the answer is zero let p+q = 1 and pq = 0.1 then log(p+ q)(p  q) = log 0.1 = 1 which fits in with log(p+q) + log(pq) = 0 + (1) = 1 because the log of 1 is 0 so when we multiply by log(p+q) we are multiplying by 0 so the log(p+q)log(p^2  q^2) = 0*1 = 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.0well I don't think this can be answered with the info given here is a counterexample to not being zero for all p and q that satisfy the condition log(p^2q^2)=1 \[\text{ Let } p+q=10^{2} \text{ and } pq=10 \\ \text{ so } p^2q^2=10^{1} \\ \text{ so } \log((p+q)(pq))=1 \\ \text{ but } \\ \log(p+q)\log(p^2q^2)=(2)(1)=2\]

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0yes  that is correct  we need more info to get a definitive answer.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0a picture would be nice as suggested by @hartnn

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but we cant assume log(p+q)log(p+q)(pq)=0, as this condition is not given. if log(p+q)(1)=? that too is in multiplication.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{ Let } p+q=10^{A} \text{ then } pq=10^{1A} \\ \text{ Since we need } \log[(p+q)(pq)]=1 \\ \text{ Verify our let part: } \\ \log[(10^{A})(10^{1A})]=\log[10^{A+(1A)}[=\log(10^{1})=1 \log(10)=1(1)=1 \\ \text{ Verified! } \\ \log(p+q)=\log(10^{A})=A \log(10)=A(1)=A \\ \log(p+q)\log(p^2q^2)=A(1)=A\] A can be any real number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and what wud be the answer in this case , 1 or 2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0Not enough info is given to apply one of the choices

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I wonder if they mean (p+q) to be the base ... if so... the problem really is: \[\log_{p+q}(pq)=1\] so we have \[(p+q)^{1}=pq\] and we are trying to find: \[\log_{p+q}((pq)(p+q)\] using product rule we have \[\log_{p+q}(pq)+\log_{p+q}(p+q)\] \[\log_{p+q}((p+q)^{1})+1\] using power rule (1)+1=0 now this is all assuming p+q is in the interval (0,1) union (1,infty)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0waooo , Great ... thanku @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.0no sorry the question doesn't make sense lol because they have more than one log in the thingy we are supposedly suppose to find

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we are not given with any base , so we have to identify by ourself that the base is given or not

freckles
 one year ago
Best ResponseYou've already chosen the best response.0there is seriously not enough information to answer this question the way it is written I have been trying to find other ways to possibly interpret the problem but i do not see a way to interpret the problem really that will lead to one of the answers but anyway I interpret it like the above I have leave something out

freckles
 one year ago
Best ResponseYou've already chosen the best response.0what are you talking about?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I didn't give an answer to your problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the above way that you have used to solve the question

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I made up a question that could have a answer that is one of your choices because the question you have could give us any of the choices (being that there is not enough information)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0answer is that, the question doesnt make any sense,. lol

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0lol if p+q is the base, then this is a really simple problem, log_(p+q) of pq = 1 log_(p+q) (p^2q^2) = log_(p+q) of (pq)(p+q) =log_(p+q) of pq + log_(p+q) of p+q = 1+1=0
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