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anonymous
 4 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 \( \log (2x)=\frac 14 \log (x15)^4 \)?
anonymous
 4 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 \( \log (2x)=\frac 14 \log (x15)^4 \)?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0q=p^2 r=p^4 s=p^6 2+4+6+1=13

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1I think too much actually :/ and act too hastily. :/ It is 13.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I like posting sitters for a change ... :P

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2I think it can have several values:\[1+4\pm2\pm6\]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2since q and s can be negative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Asnaseer, we do not know if the even number radical function is defined for strictly positive or negative. Positive is assumed.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2so are you saying, for example, that:\[s=p^6\]is not a valid solution to this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm... 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.)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2maybe FFM can clarify?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My 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.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2BTW: You are most probably correct Limitless. I am not a /true/ mathematician as such  I only follow maths as a hobby.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I hope you realize that I am still impressed by your solution, critique aside.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Asnaseer is technically right, but I am happy with the less pedantic solution for this one.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2Is there some article on this area that I could read about to help me understand this better?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0asnaseer, precisely what confuses you? This is more of a notion of convention rather than of serious mathematical concepts.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2your 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?

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1Well. 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? _

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://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. onetoone, 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. :)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2does that mean that:\[\log_p(p)^6\]is undefined?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2thx for the link Limitless.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2sorry I meant:\[\log_p(p)^6\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm.. 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 \(\log_{p}(1)\)???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean it is not defined in \( \mathbb{R}\)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2It always amazing me how even /seemingly/ simple problems can generate a whole raft of new learning! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There is a tricky problem on this topic that I encountered last month,I will post it some time later.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2BTW: The wolf says \(\log(1)=i\pi\)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2so FFM  you are right  it does not belong in the reals

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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 2dimensional 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. :)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2You just melted by brain Limitless! (in a good way) :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, I am glad I could! :) May someone more enlightened than me correct me if I am misrepresenting things.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How many values of \(x\) satisfy the equation \( \log (2x)=\frac 14 (x15)^4 \)?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2Limitless: That little plotting problem you posed up there is indeed very intriguing. Looks like my weekend is going to be filled with reading material :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Good 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well for complex numbers we need a completely new reference axes, The argand plane.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2FFM: Can this be solved analytically?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yes, 1 line if you start sufficiently from left ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This whole factor \((x15)^5\) is turning me off.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. Btw consider \(\mathbb{R} \)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2I was thinking in \(\log(2x)\), 2x cannot be negative, therefore x must be positive. And I /assumed/ that \(\frac 14 (x15)^4\) would have 2 negative and 2 positive solutions, so only 2 solutions possible.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(x\) is not an integer. Is this true?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0\( \log(2x)^4 = (x15)^4\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I missed something, then.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338633080637:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0can't find the solution analytically though!!

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2that graph makes it look like there are indeed only 2 solutions

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2symmetric in which sense?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0yes ... expoential function >> polynomial function >> log function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oops. I was wrong, sorry.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then the two solutions are x =? and x = ? .. :)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0lol ... i hope you don't mind if i use guessing technique

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2FFM: you are changing the goal posts here  your question was: How many values of x satisfy the equation...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but can't we find the solutions if there are two solutions?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i asked the same type of question of MSE they showed me Newton's method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Exact same experience. I don't even think Lambert W can be applied.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2is the log using base 10 or base e?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0common logarithm base 10 I would say.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well x=5 is the only solution. If you find the other one please let me know :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(x=5\) isn't a solution. Unless my math skills are horribly wrong right now.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2x=5 does not seem to satisfy the equation?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2this is the equation right?:\[\log (2x)=\frac 14 (x15)^4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just checked in mathematica: x = 5; 1/4 Log[10, (x  15)^4] = Log [10, 2 x]=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OMG!! I didn't mention the log carp :(

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0what am i getting?? http://www.wolframalpha.com/input/?i=solve+%28x15%29%5E4++log%5B10%2C+%282x%29%5E4%5D+%3D+0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \log (2x)=\frac 14 \log (x15)^4 \]

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2bad FFM  bad FFM  go stand in a corner now! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0VERY VERY VERY poor penman ship :( My apologies to all :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Whoa whoa whoa. One moment. Is this \(\log(x15)^4\) or is it meant to be \(\log[(x15)^4]\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think both are same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That implies \(x=15\) is a solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. that's the tricky part. Try substituting x=15 in the equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, probably not. But this is cool, so here you go: \(\log(2x)=\frac{1}{4}\log((x15)^4)\) \(4\log(2x)=\log((x15)^4)\) \(\log((2x)^4)=\log((x15)^4)\) \(2x=x15\) \(x=15\)

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2\[\log(2x)=\frac{1}{4}\log(x15)^4\]therefore:\[\log(2x)^4=\log(x15)^4\]\[(2x)^4=(x15)^4\]\[2x=\pm(x15)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes 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.

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2therefore only ONE real solution, x=5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's right asnaseer.!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Asnaseer has slain me. :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I forgot, once again, the \(\pm\).

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2:) we all have our days  in my case  seconds, of fame :)
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