Question

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

Answer 1

is it 13

Answer 2

23/12

Answer 3

13

Answer 4

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

Answer 5

I think too much actually :/ and act too hastily. :/

It is 13.

Answer 6

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

Answer 7

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

Answer 8

since q and s can be negative

Answer 9

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

Answer 10

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

Answer 11

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.)

Answer 12

maybe FFM can clarify?

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

:) thx

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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?

Answer 21

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? -_-

Answer 22

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. :)

Answer 23

good point apoorvk!

Answer 24

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

Answer 25

thx for the link Limitless.

Answer 26

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

Answer 27

In real yes.

Answer 28

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 \(\log_{p}(-1)\)???

Answer 29

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

Answer 30

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

Answer 31

*amazes

Answer 32

Yup^!

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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. :)

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

42 =_=

Answer 41

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 :)

Answer 42

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.

Answer 43

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

Answer 44

*plotting

Answer 45

FFM: Can this be solved analytically?

Answer 46

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

Answer 47

ah! - two values?

Answer 48

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

Answer 49

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

Answer 50

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.

Answer 51

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

Answer 52

No.

Answer 53

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

Answer 54

I missed something, then.