anonymous
  • anonymous
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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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
cud anybody give me the full solution. ?
anonymous
  • anonymous
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: \[\log(p+q)\log((p+q)(p-q)) = ?\]
anonymous
  • anonymous
since \[\log(p+q)(p-q) = -1\] so \[\log(p+q) -1 =0\]

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More answers

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