Poll

1 vote (16.66%)
1 vote (16.66%)
1 vote (16.66%)
1 vote (16.66%)
3 votes (50%)
1 vote (16.66%)
1 vote (16.66%)
1 vote (16.66%)
3 votes (50%)
3 votes (50%)

6 members have voted

Wizard
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January 4th, 2019 at 11:41:27 AM permalink
Solve for all possible values of x:

(x2 - 15x + 55)(x2 -9x + 20) = 1

Beer to the first correct answer and solution. As always, no searching and use spoiler tags.

The question for the poll is how many solutions are there?
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ThatDonGuy
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January 4th, 2019 at 11:55:58 AM permalink

x = 4, 5, 6, 7, 8, 9

For real numbers a and b, ab = 1 if:
(1) a = 1 -> x2 - 15x + 55 = 1 -> (x - 6)(x - 9) = 1 -> x = 6, 9
(2) b = 0 -> x2 - 9x + 20 = 0 -> (x - 4)(x - 5) = 0 -> x = 4,5
(3) a = -1 and b is even
a = -1 -> x2 - 15x + 55 = -1 -> (x - 7)(x - 8) = 0 -> x = 7,8
x = 7 -> b = 72 - 9 * 7 + 20 = 6, which is even, so x = 7 is valid
x = 8 -> b = 82 - 9 * 8 + 20 = 14, which is even, so x = 8 is valid

cmlotito
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January 4th, 2019 at 12:10:30 PM permalink
Wasn't the wall around East Berlin?
Wizard
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January 4th, 2019 at 12:27:52 PM permalink
Quote: ThatDonGuy


x = 4, 5, 6, 7, 8, 9

For real numbers a and b, ab = 1 if:
(1) a = 1 -> x2 - 15x + 55 = 1 -> (x - 6)(x - 9) = 1 -> x = 6, 9
(2) b = 0 -> x2 - 9x + 20 = 0 -> (x - 4)(x - 5) = 0 -> x = 4,5
(3) a = -1 and b is even
a = -1 -> x2 - 15x + 55 = -1 -> (x - 7)(x - 8) = 0 -> x = 7,8
x = 7 -> b = 72 - 9 * 7 + 20 = 6, which is even, so x = 7 is valid
x = 8 -> b = 82 - 9 * 8 + 20 = 14, which is even, so x = 8 is valid



Congratulations. beer++; to you.

On a related note, where is the flaw in this logic:

(-1)^2 = 1
ln(-1)^2 = ln(1)
2*ln(-1) = 0
ln(-1) = 0
pi*i = 0
exp(pi*i) = exp(0)
-1 = 1
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beachbumbabs
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January 4th, 2019 at 12:48:26 PM permalink
I think it's in this step:

ln(-1)^2 = ln(1)
2*ln(-1) = 0

The sides in the first line (above) are only equivalent because the squaring removes the - sign. 1 is unique because squaring the positive or negative of it doesn't raise the value. Any other value instead of 1 would not square to the same value. ie (-2)^2 = 4, not 2.

So, subtracting the +1 when bringing it to the left side, when exponents are a multiplication, not additive, operation leaving a net -1, is an erroneous rearrangement of the values.
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RS
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January 4th, 2019 at 1:07:05 PM permalink
Quote: Wizard

Quote: ThatDonGuy


x = 4, 5, 6, 7, 8, 9

For real numbers a and b, ab = 1 if:
(1) a = 1 -> x2 - 15x + 55 = 1 -> (x - 6)(x - 9) = 1 -> x = 6, 9
(2) b = 0 -> x2 - 9x + 20 = 0 -> (x - 4)(x - 5) = 0 -> x = 4,5
(3) a = -1 and b is even
a = -1 -> x2 - 15x + 55 = -1 -> (x - 7)(x - 8) = 0 -> x = 7,8
x = 7 -> b = 72 - 9 * 7 + 20 = 6, which is even, so x = 7 is valid
x = 8 -> b = 82 - 9 * 8 + 20 = 14, which is even, so x = 8 is valid



Congratulations. beer++; to you.

On a related note, where is the flaw in this logic:

(-1)^2 = 1
ln(-1)^2 = ln(1)
2*ln(-1) = 0
ln(-1) = 0
pi*i = 0
exp(pi*i) = exp(0)
-1 = 1


You can’t do the log of a negative number.
gamerfreak
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January 4th, 2019 at 1:14:41 PM permalink
Based on the title of this thread, I thought the answer might be 8675309
beachbumbabs
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January 4th, 2019 at 1:35:15 PM permalink
Quote: gamerfreak

Based on the title of this thread, I thought the answer might be 8675309



Ooh! So did I! I just thought it would be too weird to say so.

Wait. Yep. It is. :-p
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teliot
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January 4th, 2019 at 1:47:52 PM permalink
Quote: Wizard

ln(-1) = 0

The natural log is not a single-valued function when extended to the complex plane, so the "facts" you have used to manipulate and evaluate the log function are fake news.

Read all about it here (together with more samples of computations like yours):

http://scipp.ucsc.edu/~haber/ph116A/clog_11.pdf
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michael99000
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January 4th, 2019 at 1:54:14 PM permalink
Quote: gamerfreak

Based on the title of this thread, I thought the answer might be 8675309



It almost was
Wizard
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January 4th, 2019 at 1:55:51 PM permalink
Quote: RS

You can’t do the log of a negative number.



Sure you can.

log(-x) = log(x) + n*pi*i, where n is an odd integer.

Your homework is to prove Euler's formula: e^(bi) = cos(b) + i*sin(b)
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teliot
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January 4th, 2019 at 2:06:28 PM permalink
Quote: Wizard

Sure you can.

log(-x) = log(x) + n*pi*i, where n is an odd integer.

Your homework is to prove Euler's formula: e^(bi) = cos(b) + i*sin(b)

It's not where n is "an" odd integer, it is where n is "all" odd integers. You can choose a particular region to define it for convenience, but when/if it wraps, your computation will be foiled.

https://en.wikipedia.org/wiki/Riemann_surface#/media/File:Riemann_surface_log.svg

I think this is a very esoteric point about Riemann surfaces on which to base a conundrum such as the one you posted.
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billryan
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January 4th, 2019 at 2:31:00 PM permalink
Sadly, I'd understand this thread equally well if the whole thing was in Swahili.
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onenickelmiracle
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January 4th, 2019 at 2:33:28 PM permalink
Quote: gamerfreak

Based on the title of this thread, I thought the answer might be 8675309

Or x=$200
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ThatDonGuy
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January 4th, 2019 at 3:25:32 PM permalink
Quote: Wizard

Sure you can.

log(-x) = log(x) + n*pi*i, where n is an odd integer.

Your homework is to prove Euler's formula: e^(bi) = cos(b) + i*sin(b)



Use Taylor's Theorem (don't ask me to prove that...)

The nth derivative of ex with respect to x is ex
By Taylor's theorem (with respect to x = 0), ebi = 1 + bi + (bi)2 / 2! + (bi)3 / 3! + ...
= (1 + b2i2 / 2! + b4i4 / 4! + b6i6 / 6!+ ...)
+ (bi / 1! + b3i3 / 3! + b5i5 / 5! + b7i7 / 7! + ...)
= (1 - b2 / 2! + b4 / 4! - b6 / 6! + ...) + i (b - b3 / 3! + b5 / 5! - b7 / 7! + ...)
Again by Taylor's theorem, cos b (with respect to 0) = 1 - b2 / 2! + b4 / 4! - b6 / 6! + ...,
and sin b (with respect to 0) = b - b3 / 3! + b5 / 5! - b7 / 7! + ...),
which means cos b + i sin b = (1 - b2 / 2! + b4 / 4! - b6 / 6! + ...) + i (b - b3 / 3! + b5 / 5! - b7 / 7! + ...) = ebi

Wizard
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January 4th, 2019 at 3:47:57 PM permalink
Quote: ThatDonGuy


Use Taylor's Theorem (don't ask me to prove that...)

The nth derivative of ex with respect to x is ex
By Taylor's theorem (with respect to x = 0), ebi = 1 + bi + (bi)2 / 2! + (bi)3 / 3! + ...
= (1 + b2i2 / 2! + b4i4 / 4! + b6i6 / 6!+ ...)
+ (bi / 1! + b3i3 / 3! + b5i5 / 5! + b7i7 / 7! + ...)
= (1 - b2 / 2! + b4 / 4! - b6 / 6! + ...) + i (b - b3 / 3! + b5 / 5! - b7 / 7! + ...)
Again by Taylor's theorem, cos b (with respect to 0) = 1 - b2 / 2! + b4 / 4! - b6 / 6! + ...,
and sin b (with respect to 0) = b - b3 / 3! + b5 / 5! - b7 / 7! + ...),
which means cos b + i sin b = (1 - b2 / 2! + b4 / 4! - b6 / 6! + ...) + i (b - b3 / 3! + b5 / 5! - b7 / 7! + ...) = ebi



Very good. Actually, I meant the homework assignment for RS, but I doubt very highly he would have even started on it. Hopefully he will carefully go over your answer before claiming you can't take the log of a negative number.

New question: What is ln(degeneres)?
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ThatDonGuy
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January 4th, 2019 at 4:23:40 PM permalink
Quote: Wizard

New question: What is ln(degeneres)?


Er, is calling an openly LGBTQ person "what" a moderatable offense?

Oh, and if we can't get a Spring Fling together, I have June 16-20 penciled in for Vegas.
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January 4th, 2019 at 5:02:32 PM permalink
Quote: billryan

Sadly, I'd understand this thread equally well if the whole thing was in Swahili.

Kazi yako ya nyumbani ni kuthibitisha formula ya Euler: e ^ (bi) = cos (b) + i * sin (b)
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Wizard
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January 4th, 2019 at 5:07:49 PM permalink
Quote: ThatDonGuy

Er, is calling an openly LGBTQ person "what" a moderatable offense?



Hmmm. I didn't think of it that way. How about this wording: solve for x where x=ln(degeneres).
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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