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

- anonymous

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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- anonymous

cud anybody give me the full solution. ?

- 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

since \[\log(p+q)(p-q) = -1\]
so \[\log(p+q) -1 =0\]

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## More answers

- 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

@welshfella that's in my last reply

- anonymous

log(p+q) = 1

- anonymous

so since log(p+q) = 1
it will be:
\[\log(p+q)\log(p^2-q^2) = 0\]

- welshfella

i cant see your logic there - I'm not saying you are wrong

- 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

if log ( p + q) = 1
log(p+q)log(p^-q^2) = 1 * -1 = -1 not 0

- anonymous

oh yes my bad I was subtracting
then yes it would be -1

- freckles

@nopen it looked like you said:
\[\log(a \cdot b)=\log(a) \cdot \log(b) \text{ thisn't a log property though }\]

- hartnn

can the asker post the screenshot of the question?

- anonymous

but isn't it
log(a*b) = log(a) + log(b)?

- freckles

yes that is true

- freckles

that is the product rule for log

- 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

yeah its d

- 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

yes - that is correct - we need more info to get a definitive answer.

- freckles

a picture would be nice as suggested by @hartnn

- welshfella

yes

- anonymous

thank you @nopen

- 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

##### 1 Attachment

- 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

and what wud be the answer in this case , 1 or 2?

- freckles

no way to tell

- freckles

Not enough info is given to apply one of the choices

- 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

waooo , Great ... thanku @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

but it seems good

- anonymous

we are not given with any base , so we have to identify by ourself that the base is given or not

- 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

that is not ryt

- freckles

what are you talking about?

- freckles

the answer?

- freckles

I didn't give an answer to your problem

- anonymous

the above way that you have used to solve the question

- anonymous

taking base p+q

- 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

fine

- anonymous

answer is that, the question doesnt make any sense,. lol

- 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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