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

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is it 13

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Other answers:

q=p^2 r=p^4 s=p^6 2+4+6+1=13
I think too much actually :/ and act too hastily. :/ It is 13.
I like posting sitters for a change ... :P
I think it can have several values:\[1+4\pm2\pm6\]
since q and s can be negative
Asnaseer, we do not know if the even number radical function is defined for strictly positive or negative. Positive is assumed.
so are you saying, for example, that:\[s=-p^6\]is not a valid solution to this?
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.)
maybe FFM can clarify?
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.
BTW: You are most probably correct Limitless. I am not a /true/ mathematician as such - I only follow maths as a hobby.
I hope you realize that I am still impressed by your solution, critique aside.
:) thx
Asnaseer is technically right, but I am happy with the less pedantic solution for this one.
Is there some article on this area that I could read about to help me understand this better?
asnaseer, precisely what confuses you? This is more of a notion of convention rather than of serious mathematical concepts.
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?
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? -_- :) 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. :)
good point apoorvk!
does that mean that:\[\log_p(-p)^6\]is undefined?
thx for the link Limitless.
sorry I meant:\[\log_p-(p)^6\]
In real yes.
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 \(\log_{p}(-1)\)???
I mean it is not defined in \( \mathbb{R}\)
It always amazing me how even /seemingly/ simple problems can generate a whole raft of new learning! :)
There is a tricky problem on this topic that I encountered last month,I will post it some time later.
BTW: The wolf says \(\log(-1)=i\pi\)
so FFM - you are right - it does not belong in the reals
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. :)
You just melted by brain Limitless! (in a good way) :D
Well, I am glad I could! :) May someone more enlightened than me correct me if I am misrepresenting things.
How many values of \(x\) satisfy the equation \( \log (2x)=\frac 14 (x-15)^4 \)?
42 =_=
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 :)
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.
Well for complex numbers we need a completely new reference axes, The argand plane.
FFM: Can this be solved analytically?
Oh yes, 1 line if you start sufficiently from left ;)
ah! - two values?
This whole factor \((x-15)^5\) is turning me off.
No. Btw consider \(\mathbb{R} \)
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.
\(x\) is not an integer. Is this true?
\( \log(2x)^4 = (x-15)^4\)
I missed something, then.
can't find the solution analytically though!!
that graph makes it look like there are indeed only 2 solutions
Which are symmetric.
symmetric in which sense?
yes ... expoential function >> polynomial function >> log function
Oops. I was wrong, sorry.
Then the two solutions are x =? and x = ? .. :)
lol ... i hope you don't mind if i use guessing technique
FFM: you are changing the goal posts here - your question was: How many values of x satisfy the equation...
Yes, but can't we find the solutions if there are two solutions?
i asked the same type of question of MSE they showed me Newton's method
lol :D
Exact same experience. I don't even think Lambert W can be applied.
is the log using base 10 or base e?
common logarithm base 10 I would say.
Well x=5 is the only solution. If you find the other one please let me know :)
\(x=5\) isn't a solution. Unless my math skills are horribly wrong right now.
x=5 does not seem to satisfy the equation?
this is the equation right?:\[\log (2x)=\frac 14 (x-15)^4\]
I just checked in mathematica: x = 5; 1/4 Log[10, (x - 15)^4] = Log [10, 2 x]=1
OMG!! I didn't mention the log carp :(
what am i getting??
\[ \log (2x)=\frac 14 \log (x-15)^4 \]
bad FFM - bad FFM - go stand in a corner now! :)
VERY VERY VERY poor penman ship :( My apologies to all :(
Whoa whoa whoa. One moment. Is this \(\log(x-15)^4\) or is it meant to be \(\log[(x-15)^4]\)?
I think both are same.
That implies \(x=-15\) is a solution.
No. that's the tricky part. Try substituting x=-15 in the equation.
So, probably not. But this is cool, so here you go: \(\log(2x)=\frac{1}{4}\log((x-15)^4)\) \(4\log(2x)=\log((x-15)^4)\) \(\log((2x)^4)=\log((x-15)^4)\) \(2x=x-15\) \(x=-15\)
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.
therefore only ONE real solution, x=5
That's right asnaseer.!
Asnaseer has slain me. :p
I forgot, once again, the \(\pm\).
:) we all have our days - in my case - seconds, of fame :)

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