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FoolForMath
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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 \( \log (2x)=\frac 14 \log (x15)^4 \)?
 2 years ago
 2 years ago
FoolForMath Group Title
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 \( \log (2x)=\frac 14 \log (x15)^4 \)?
 2 years ago
 2 years ago

This Question is Closed

Limitless Group TitleBest ResponseYou've already chosen the best response.2
q=p^2 r=p^4 s=p^6 2+4+6+1=13
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I like posting sitters for a change ... :P
 2 years ago

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

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Asnaseer, we do not know if the even number radical function is defined for strictly positive or negative. Positive is assumed.
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
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.)
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
I hope you realize that I am still impressed by your solution, critique aside.
 2 years ago

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

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
asnaseer, precisely what confuses you? This is more of a notion of convention rather than of serious mathematical concepts.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
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?
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.1
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? _
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
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. 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. :)
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
good point apoorvk!
 2 years ago

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

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

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
In real yes.
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
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 \(\log_{p}(1)\)???
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I mean it is not defined in \( \mathbb{R}\)
 2 years ago

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

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

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

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
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 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. :)
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Well, I am glad I could! :) May someone more enlightened than me correct me if I am misrepresenting things.
 2 years ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
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 :)
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Well for complex numbers we need a completely new reference axes, The argand plane.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
*plotting
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Oh yes, 1 line if you start sufficiently from left ;)
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
ah!  two values?
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
This whole factor \((x15)^5\) is turning me off.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
No. Btw consider \(\mathbb{R} \)
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
I 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.
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
\(x\) is not an integer. Is this true?
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
I missed something, then.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1338633080637:dw
 2 years ago

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

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Which are symmetric.
 2 years ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yes ... expoential function >> polynomial function >> log function
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Oops. I was wrong, sorry.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Then the two solutions are x =? and x = ? .. :)
 2 years ago

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

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
Yes, but can't we find the solutions if there are two solutions?
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Exact same experience. I don't even think Lambert W can be applied.
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
common logarithm base 10 I would say.
 2 years ago

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

Limitless Group TitleBest ResponseYou've already chosen the best response.2
\(x=5\) isn't a solution. Unless my math skills are horribly wrong right now.
 2 years ago

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

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

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
OMG!! I didn't mention the log carp :(
 2 years ago

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

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

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
VERY VERY VERY poor penman ship :( My apologies to all :(
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Whoa whoa whoa. One moment. Is this \(\log(x15)^4\) or is it meant to be \(\log[(x15)^4]\)?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
I think both are same.
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
That implies \(x=15\) is a solution.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
No. that's the tricky part. Try substituting x=15 in the equation.
 2 years ago

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

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.0
That's right asnaseer.!
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
Asnaseer has slain me. :p
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.2
I forgot, once again, the \(\pm\).
 2 years ago

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