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

- anonymous

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- anonymous

is it 13

- apoorvk

23/12

- apoorvk

13

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## More answers

- anonymous

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

- apoorvk

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

- anonymous

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

- asnaseer

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

- asnaseer

since q and s can be negative

- anonymous

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

- asnaseer

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

- 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

maybe FFM can clarify?

- 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

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

- anonymous

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

- asnaseer

:) thx

- anonymous

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

- asnaseer

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

- anonymous

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

- 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

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

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

good point apoorvk!

- asnaseer

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

- asnaseer

thx for the link Limitless.

- asnaseer

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

- anonymous

In real yes.

- 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

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

- asnaseer

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

- asnaseer

*amazes

- apoorvk

Yup^!

- anonymous

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

- asnaseer

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

- asnaseer

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

- 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

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

- anonymous

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

- anonymous

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

- lgbasallote

42 =_=

- 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

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

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

- anonymous

*plotting

- asnaseer

FFM: Can this be solved analytically?

- anonymous

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

- asnaseer

ah! - two values?

- anonymous

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

- anonymous

No. Btw consider \(\mathbb{R} \)

- 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

\(x\) is not an integer. Is this true?

- anonymous

No.

- experimentX

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

- anonymous

I missed something, then.

- experimentX

|dw:1338633080637:dw|

- experimentX

can't find the solution analytically though!!

- asnaseer

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

- anonymous

Which are symmetric.

- asnaseer

symmetric in which sense?

- experimentX

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

- anonymous

Oops. I was wrong, sorry.

- anonymous

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

- experimentX

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

- asnaseer

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

- anonymous

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

- experimentX

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

- experimentX

lol :D

- anonymous

^yes.

- anonymous

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

- asnaseer

is the log using base 10 or base e?

- anonymous

common logarithm base 10 I would say.

- anonymous

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

- anonymous

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

- asnaseer

x=5 does not seem to satisfy the equation?

- asnaseer

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

- anonymous

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

- anonymous

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

- 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

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

- asnaseer

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

- anonymous

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

- anonymous

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

- anonymous

I think both are same.

- anonymous

That implies \(x=-15\) is a solution.

- anonymous

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

- 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

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

- anonymous

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

therefore only ONE real solution, x=5

- anonymous

That's right asnaseer.!

- anonymous

Asnaseer has slain me. :p

- anonymous

I forgot, once again, the \(\pm\).

- asnaseer

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

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