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is it 13

23/12

13

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

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

maybe FFM can clarify?

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?

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.

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

*amazes

Yup^!

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

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

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

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

42 =_=

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

*plotting

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 \(\mathbb{R} \)

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

No.

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

I missed something, then.

|dw:1338633080637:dw|

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

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

^yes.

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 :(

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

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.

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 :)