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

1. anonymous

cud anybody give me the full solution. ?

2. 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: $\log(p+q)\log((p+q)(p-q)) = ?$

3. anonymous

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

4. 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.

5. anonymous

@welshfella that's in my last reply

6. anonymous

log(p+q) = 1

7. anonymous

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

8. welshfella

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

9. 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

10. welshfella

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

11. anonymous

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

12. freckles

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

13. hartnn

can the asker post the screenshot of the question?

14. anonymous

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

15. freckles

yes that is true

16. freckles

that is the product rule for log

17. 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

18. welshfella

yeah its d

19. 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 $\text{ Let } p+q=10^{-2} \text{ and } p-q=10 \\ \text{ so } p^2-q^2=10^{-1} \\ \text{ so } \log((p+q)(p-q))=-1 \\ \text{ but } \\ \log(p+q)\log(p^2-q^2)=(-2)(-1)=2$

20. welshfella

21. freckles

a picture would be nice as suggested by @hartnn

22. welshfella

yes

23. anonymous

thank you @nopen

24. 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.

25. anonymous

26. 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

27. anonymous

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

28. freckles

no way to tell

29. freckles

Not enough info is given to apply one of the choices

30. 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)

31. anonymous

waooo , Great ... thanku @freckles

32. 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

33. anonymous

but it seems good

34. anonymous

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

35. 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

36. anonymous

that is not ryt

37. freckles

38. freckles

39. freckles

40. anonymous

the above way that you have used to solve the question

41. anonymous

taking base p+q

42. 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)

43. anonymous

fine

44. anonymous

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

45. 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