## FoolForMath 3 years ago Easy yet cute problem, if $$p^{12}=q^6=r^3=s^2, p \neq 1$$. Find the value of $$\log_p pqrs$$. Lets see who can solve this the fastest. Pleas read on for the extra question (posted later): How many values of $$x$$ satisfy the equation $$\log (2x)=\frac 14 \log (x-15)^4$$?

1. matricked

is it 13

2. apoorvk

23/12

3. apoorvk

13

4. Limitless

q=p^2 r=p^4 s=p^6 2+4+6+1=13

5. apoorvk

I think too much actually :/ and act too hastily. :/ It is 13.

6. FoolForMath

I like posting sitters for a change ... :P

7. asnaseer

I think it can have several values:$1+4\pm2\pm6$

8. asnaseer

since q and s can be negative

9. Limitless

Asnaseer, we do not know if the even number radical function is defined for strictly positive or negative. Positive is assumed.

10. asnaseer

so are you saying, for example, that:$s=-p^6$is not a valid solution to this?

11. Limitless

Hmm... Rather hard to say exactly what I mean. It's true that $$s=-p^6$$. But, I assumed we were using the positive roots only because that's conventional. Otherwise, as you see, our function is actually a multivalued relation rather than a function and thus the logarithm becomes murkily defined. (I am potentially abusing terminology here.)

12. asnaseer

maybe FFM can clarify?

13. Limitless

My main point is that if we allow ourselves to take both of the roots, the logarithm is no longer a function. This can't be the case, since the logarithm is always a function.

14. asnaseer

BTW: You are most probably correct Limitless. I am not a /true/ mathematician as such - I only follow maths as a hobby.

15. Limitless

I hope you realize that I am still impressed by your solution, critique aside.

16. asnaseer

:) thx

17. FoolForMath

Asnaseer is technically right, but I am happy with the less pedantic solution for this one.

18. asnaseer

Is there some article on this area that I could read about to help me understand this better?

19. Limitless

asnaseer, precisely what confuses you? This is more of a notion of convention rather than of serious mathematical concepts.

20. asnaseer

your description of log being a function and therefore it can only have one value - I was just wondering if there was any formal definition of this concept somewhere?

21. apoorvk

Well. I seem to think that 'q' and 's' can be negative, but their resultant powers would definitely be positive - so how does the question of the powers being 'negative' arise? s= (-p)^6 ---> alrighty! but the exponent can only be '6' right? of the exponent changes sign, then the whole thingy changes. Or am I totally wrong once again? -_-

22. Limitless

http://en.wikipedia.org/wiki/Function_(mathematics)#Formal_definition :) I rather love formal definitions if they are also intuitively beautiful. This is one such case. The reason that functions are considered injective and surjective (i.e. one-to-one, which is what I have been saying) is that this allows us to suppose the existence of their inversion via another function! It's rather interesting stuff. :)

23. asnaseer

good point apoorvk!

24. asnaseer

does that mean that:$\log_p(-p)^6$is undefined?

25. asnaseer

26. asnaseer

sorry I meant:$\log_p-(p)^6$

27. FoolForMath

In real yes.

28. Limitless

hmm.. Good question. $$\log_{p}(-p^6)=\log_{p}(-1)+\log_{p}(p^6)=\log_{p}(-1)+6$$ Here's an intuitive way of thinking: What in the world is $$\log_{p}(-1)$$???

29. FoolForMath

I mean it is not defined in $$\mathbb{R}$$

30. asnaseer

It always amazing me how even /seemingly/ simple problems can generate a whole raft of new learning! :)

31. asnaseer

*amazes

32. apoorvk

Yup^!

33. FoolForMath

There is a tricky problem on this topic that I encountered last month,I will post it some time later.

34. asnaseer

BTW: The wolf says $$\log(-1)=i\pi$$

35. asnaseer

so FFM - you are right - it does not belong in the reals

36. Limitless

You will find it very intriguing if you ask yourself why we restrict ourselves to certain domains on popular function. It taught me many deep ideas to ponder this. A rather elementary reason is that much of a function cannot even be graphed on a real 2-dimensional plane. Consider $$f(x)=x^2$$ at $$x=i$$. It is clear that we can graph $$f(i)=-1$$, however how do we plot $$i$$??? This is where it becomes very intriguing: Complex numbers require special plotting techniques or special geometric spaces. :)

37. asnaseer

You just melted by brain Limitless! (in a good way) :D

38. Limitless

Well, I am glad I could! :) May someone more enlightened than me correct me if I am misrepresenting things.

39. FoolForMath

How many values of $$x$$ satisfy the equation $$\log (2x)=\frac 14 (x-15)^4$$?

40. lgbasallote

42 =_=

41. asnaseer

Limitless: That little plotting problem you posed up there is indeed very intriguing. Looks like my weekend is going to be filled with reading material :)

42. Limitless

Good lord, FFM. Surely this does not require numerical methods. D: I hope so, Asn. I had much spare time in class, LOL. You ask yourself questions like this out of boredom.

43. FoolForMath

Well for complex numbers we need a completely new reference axes, The argand plane.

44. FoolForMath

*plotting

45. asnaseer

FFM: Can this be solved analytically?

46. FoolForMath

Oh yes, 1 line if you start sufficiently from left ;)

47. asnaseer

ah! - two values?

48. Limitless

This whole factor $$(x-15)^5$$ is turning me off.

49. FoolForMath

No. Btw consider $$\mathbb{R}$$

50. asnaseer

I was thinking in $$\log(2x)$$, 2x cannot be negative, therefore x must be positive. And I /assumed/ that $$\frac 14 (x-15)^4$$ would have 2 negative and 2 positive solutions, so only 2 solutions possible.

51. Limitless

$$x$$ is not an integer. Is this true?

52. FoolForMath

No.

53. experimentX

$$\log(2x)^4 = (x-15)^4$$

54. Limitless

I missed something, then.

55. experimentX

|dw:1338633080637:dw|

56. experimentX

can't find the solution analytically though!!

57. asnaseer

that graph makes it look like there are indeed only 2 solutions

58. Limitless

Which are symmetric.

59. asnaseer

symmetric in which sense?

60. experimentX

yes ... expoential function >> polynomial function >> log function

61. Limitless

Oops. I was wrong, sorry.

62. FoolForMath

Then the two solutions are x =? and x = ? .. :)

63. experimentX

lol ... i hope you don't mind if i use guessing technique

64. asnaseer

FFM: you are changing the goal posts here - your question was: How many values of x satisfy the equation...

65. FoolForMath

Yes, but can't we find the solutions if there are two solutions?

66. experimentX

i asked the same type of question of MSE they showed me Newton's method

67. experimentX

lol :D

68. Limitless

^yes.

69. Limitless

Exact same experience. I don't even think Lambert W can be applied.

70. asnaseer

is the log using base 10 or base e?

71. FoolForMath

common logarithm base 10 I would say.

72. FoolForMath

Well x=5 is the only solution. If you find the other one please let me know :)

73. Limitless

$$x=5$$ isn't a solution. Unless my math skills are horribly wrong right now.

74. asnaseer

x=5 does not seem to satisfy the equation?

75. asnaseer

this is the equation right?:$\log (2x)=\frac 14 (x-15)^4$

76. FoolForMath

I just checked in mathematica: x = 5; 1/4 Log[10, (x - 15)^4] = Log [10, 2 x]=1

77. FoolForMath

OMG!! I didn't mention the log carp :(

78. experimentX
79. FoolForMath

$\log (2x)=\frac 14 \log (x-15)^4$

80. asnaseer

bad FFM - bad FFM - go stand in a corner now! :)

81. FoolForMath

VERY VERY VERY poor penman ship :( My apologies to all :(

82. Limitless

Whoa whoa whoa. One moment. Is this $$\log(x-15)^4$$ or is it meant to be $$\log[(x-15)^4]$$?

83. FoolForMath

I think both are same.

84. Limitless

That implies $$x=-15$$ is a solution.

85. FoolForMath

No. that's the tricky part. Try substituting x=-15 in the equation.

86. Limitless

So, probably not. But this is cool, so here you go: $$\log(2x)=\frac{1}{4}\log((x-15)^4)$$ $$4\log(2x)=\log((x-15)^4)$$ $$\log((2x)^4)=\log((x-15)^4)$$ $$2x=x-15$$ $$x=-15$$

87. asnaseer

$\log(2x)=\frac{1}{4}\log(x-15)^4$therefore:$\log(2x)^4=\log(x-15)^4$$(2x)^4=(x-15)^4$$2x=\pm(x-15)$

88. FoolForMath

Yes yes but as I said x=-15 is not a valid solution. Okay I am going to going to dig a hole and hide my face on it. I will see you guys later.

89. asnaseer

therefore only ONE real solution, x=5

90. FoolForMath

That's right asnaseer.!

91. Limitless

Asnaseer has slain me. :p

92. Limitless

I forgot, once again, the $$\pm$$.

93. asnaseer

:) we all have our days - in my case - seconds, of fame :)