anonymous
  • anonymous
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 \( \log (2x)=\frac 14 \log (x-15)^4 \)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
is it 13
apoorvk
  • apoorvk
23/12
apoorvk
  • apoorvk
13

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anonymous
  • anonymous
q=p^2 r=p^4 s=p^6 2+4+6+1=13
apoorvk
  • apoorvk
I think too much actually :/ and act too hastily. :/ It is 13.
anonymous
  • anonymous
I like posting sitters for a change ... :P
asnaseer
  • asnaseer
I think it can have several values:\[1+4\pm2\pm6\]
asnaseer
  • asnaseer
since q and s can be negative
anonymous
  • anonymous
Asnaseer, we do not know if the even number radical function is defined for strictly positive or negative. Positive is assumed.
asnaseer
  • asnaseer
so are you saying, for example, that:\[s=-p^6\]is not a valid solution to this?
anonymous
  • anonymous
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.)
asnaseer
  • asnaseer
maybe FFM can clarify?
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
BTW: You are most probably correct Limitless. I am not a /true/ mathematician as such - I only follow maths as a hobby.
anonymous
  • anonymous
I hope you realize that I am still impressed by your solution, critique aside.
asnaseer
  • asnaseer
:) thx
anonymous
  • anonymous
Asnaseer is technically right, but I am happy with the less pedantic solution for this one.
asnaseer
  • asnaseer
Is there some article on this area that I could read about to help me understand this better?
anonymous
  • anonymous
asnaseer, precisely what confuses you? This is more of a notion of convention rather than of serious mathematical concepts.
asnaseer
  • 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?
apoorvk
  • 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? -_-
anonymous
  • anonymous
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. :)
asnaseer
  • asnaseer
good point apoorvk!
asnaseer
  • asnaseer
does that mean that:\[\log_p(-p)^6\]is undefined?
asnaseer
  • asnaseer
thx for the link Limitless.
asnaseer
  • asnaseer
sorry I meant:\[\log_p-(p)^6\]
anonymous
  • anonymous
In real yes.
anonymous
  • anonymous
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 \(\log_{p}(-1)\)???
anonymous
  • anonymous
I mean it is not defined in \( \mathbb{R}\)
asnaseer
  • asnaseer
It always amazing me how even /seemingly/ simple problems can generate a whole raft of new learning! :)
asnaseer
  • asnaseer
*amazes
apoorvk
  • apoorvk
Yup^!
anonymous
  • anonymous
There is a tricky problem on this topic that I encountered last month,I will post it some time later.
asnaseer
  • asnaseer
BTW: The wolf says \(\log(-1)=i\pi\)
asnaseer
  • asnaseer
so FFM - you are right - it does not belong in the reals
anonymous
  • anonymous
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. :)
asnaseer
  • asnaseer
You just melted by brain Limitless! (in a good way) :D
anonymous
  • anonymous
Well, I am glad I could! :) May someone more enlightened than me correct me if I am misrepresenting things.
anonymous
  • anonymous
How many values of \(x\) satisfy the equation \( \log (2x)=\frac 14 (x-15)^4 \)?
lgbasallote
  • lgbasallote
42 =_=
asnaseer
  • 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 :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Well for complex numbers we need a completely new reference axes, The argand plane.
anonymous
  • anonymous
*plotting
asnaseer
  • asnaseer
FFM: Can this be solved analytically?
anonymous
  • anonymous
Oh yes, 1 line if you start sufficiently from left ;)
asnaseer
  • asnaseer
ah! - two values?
anonymous
  • anonymous
This whole factor \((x-15)^5\) is turning me off.
anonymous
  • anonymous
No. Btw consider \(\mathbb{R} \)
asnaseer
  • 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.
anonymous
  • anonymous
\(x\) is not an integer. Is this true?
anonymous
  • anonymous
No.
experimentX
  • experimentX
\( \log(2x)^4 = (x-15)^4\)
anonymous
  • anonymous
I missed something, then.
experimentX
  • experimentX
|dw:1338633080637:dw|
experimentX
  • experimentX
can't find the solution analytically though!!
asnaseer
  • asnaseer
that graph makes it look like there are indeed only 2 solutions
anonymous
  • anonymous
Which are symmetric.
asnaseer
  • asnaseer
symmetric in which sense?
experimentX
  • experimentX
yes ... expoential function >> polynomial function >> log function
anonymous
  • anonymous
Oops. I was wrong, sorry.
anonymous
  • anonymous
Then the two solutions are x =? and x = ? .. :)
experimentX
  • experimentX
lol ... i hope you don't mind if i use guessing technique
asnaseer
  • asnaseer
FFM: you are changing the goal posts here - your question was: How many values of x satisfy the equation...
anonymous
  • anonymous
Yes, but can't we find the solutions if there are two solutions?
experimentX
  • experimentX
i asked the same type of question of MSE they showed me Newton's method
experimentX
  • experimentX
lol :D
anonymous
  • anonymous
^yes.
anonymous
  • anonymous
Exact same experience. I don't even think Lambert W can be applied.
asnaseer
  • asnaseer
is the log using base 10 or base e?
anonymous
  • anonymous
common logarithm base 10 I would say.
anonymous
  • anonymous
Well x=5 is the only solution. If you find the other one please let me know :)
anonymous
  • anonymous
\(x=5\) isn't a solution. Unless my math skills are horribly wrong right now.
asnaseer
  • asnaseer
x=5 does not seem to satisfy the equation?
asnaseer
  • asnaseer
this is the equation right?:\[\log (2x)=\frac 14 (x-15)^4\]
anonymous
  • anonymous
I just checked in mathematica: x = 5; 1/4 Log[10, (x - 15)^4] = Log [10, 2 x]=1
anonymous
  • anonymous
OMG!! I didn't mention the log carp :(
experimentX
  • experimentX
what am i getting?? http://www.wolframalpha.com/input/?i=solve+%28x-15%29%5E4+-+log%5B10%2C+%282x%29%5E4%5D+%3D+0
anonymous
  • anonymous
\[ \log (2x)=\frac 14 \log (x-15)^4 \]
asnaseer
  • asnaseer
bad FFM - bad FFM - go stand in a corner now! :)
anonymous
  • anonymous
VERY VERY VERY poor penman ship :( My apologies to all :(
anonymous
  • anonymous
Whoa whoa whoa. One moment. Is this \(\log(x-15)^4\) or is it meant to be \(\log[(x-15)^4]\)?
anonymous
  • anonymous
I think both are same.
anonymous
  • anonymous
That implies \(x=-15\) is a solution.
anonymous
  • anonymous
No. that's the tricky part. Try substituting x=-15 in the equation.
anonymous
  • anonymous
So, probably not. But this is cool, so here you go: \(\log(2x)=\frac{1}{4}\log((x-15)^4)\) \(4\log(2x)=\log((x-15)^4)\) \(\log((2x)^4)=\log((x-15)^4)\) \(2x=x-15\) \(x=-15\)
asnaseer
  • asnaseer
\[\log(2x)=\frac{1}{4}\log(x-15)^4\]therefore:\[\log(2x)^4=\log(x-15)^4\]\[(2x)^4=(x-15)^4\]\[2x=\pm(x-15)\]
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
therefore only ONE real solution, x=5
anonymous
  • anonymous
That's right asnaseer.!
anonymous
  • anonymous
Asnaseer has slain me. :p
anonymous
  • anonymous
I forgot, once again, the \(\pm\).
asnaseer
  • asnaseer
:) we all have our days - in my case - seconds, of fame :)

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