log(p+q)(p-q) = -1 log(p+q)log(p^2-q^2) = ? option: a)2 b)1 c) -1 d) 0

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log(p+q)(p-q) = -1 log(p+q)log(p^2-q^2) = ? option: a)2 b)1 c) -1 d) 0

Linear Algebra
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cud anybody give me the full solution. ?
umm I don't know if this is correct: \[\log(p+q)\log(p^2-q^2) =?\] \[\log(p+q)\log(p-q) = -1\] so first equation is turned into: \[\log(p+q)\log((p+q)(p-q)) = ?\]
since \[\log(p+q)(p-q) = -1\] so \[\log(p+q) -1 =0\]

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log(p^2-q^2) = log(p+q)(p-q) = -1 but i cant see a way to find what log(p+q) is.
@welshfella that's in my last reply
log(p+q) = 1
so since log(p+q) = 1 it will be: \[\log(p+q)\log(p^2-q^2) = 0\]
i cant see your logic there - I'm not saying you are wrong
I mean to say that since log(p^2-q^2) = -1 log(p+q)log(p^2-q^2) = will be log(p+q) -1 = 0 so log(p+q) = 1
if log ( p + q) = 1 log(p+q)log(p^-q^2) = 1 * -1 = -1 not 0
oh yes my bad I was subtracting then yes it would be -1
@nopen it looked like you said: \[\log(a \cdot b)=\log(a) \cdot \log(b) \text{ thisn't a log property though }\]
can the asker post the screenshot of the question?
but isn't it log(a*b) = log(a) + log(b)?
yes that is true
that is the product rule for log
I think the answer is zero let p+q = 1 and p-q = 0.1 then log(p+ q)(p - q) = log 0.1 = -1 which fits in with log(p+q) + log(p-q) = 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
yeah its d
well 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^2-q^2)=-1 \[\text{ Let } p+q=10^{-2} \text{ and } p-q=10 \\ \text{ so } p^2-q^2=10^{-1} \\ \text{ so } \log((p+q)(p-q))=-1 \\ \text{ but } \\ \log(p+q)\log(p^2-q^2)=(-2)(-1)=2\]
yes - that is correct - we need more info to get a definitive answer.
a picture would be nice as suggested by @hartnn
yes
thank you @nopen
but we cant assume log(p+q)log(p+q)(p-q)=0, as this condition is not given. if log(p+q)(-1)=? that too is in multiplication.
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\[\text{ Let } p+q=10^{A} \text{ then } p-q=10^{-1-A} \\ \text{ Since we need } \log[(p+q)(p-q)]=-1 \\ \text{ Verify our let part: } \\ \log[(10^{A})(10^{-1-A})]=\log[10^{A+(-1-A)}[=\log(10^{-1})=-1 \log(10)=-1(1)=-1 \\ \text{ Verified! } \\ \log(p+q)=\log(10^{A})=A \log(10)=A(1)=A \\ \log(p+q)\log(p^2-q^2)=A(-1)=-A\] A can be any real number
and what wud be the answer in this case , 1 or 2?
no way to tell
Not enough info is given to apply one of the choices
I wonder if they mean (p+q) to be the base ... if so... the problem really is: \[\log_{p+q}(p-q)=-1\] so we have \[(p+q)^{-1}=p-q\] and we are trying to find: \[\log_{p+q}((p-q)(p+q)\] using product rule we have \[\log_{p+q}(p-q)+\log_{p+q}(p+q)\] \[\log_{p+q}((p+q)^{-1})+1\] using power rule (-1)+1=0 now this is all assuming p+q is in the interval (0,1) union (1,infty)
waooo , Great ... thanku @freckles
no sorry the question doesn't make sense lol because they have more than one log in the thingy we are supposedly suppose to find
but it seems good
we are not given with any base , so we have to identify by ourself that the base is given or not
there 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
that is not ryt
what are you talking about?
the answer?
I didn't give an answer to your problem
the above way that you have used to solve the question
taking base p+q
I 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)
fine
answer is that, the question doesnt make any sense,. lol
lol if p+q is the base, then this is a really simple problem, log_(p+q) of p-q = -1 log_(p+q) (p^2-q^2) = log_(p+q) of (p-q)(p+q) =log_(p+q) of p-q + log_(p+q) of p+q = -1+1=0

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