For a good time, solve this equation
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6 members have voted
January 4th, 2019 at 11:41:27 AM
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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?
(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?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 4th, 2019 at 11:55:58 AM
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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
January 4th, 2019 at 12:10:30 PM
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Wasn't the wall around East Berlin?
January 4th, 2019 at 12:27:52 PM
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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
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 4th, 2019 at 12:48:26 PM
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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.
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.
If the House lost every hand, they wouldn't deal the game.
January 4th, 2019 at 1:07:05 PM
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Quote: WizardQuote: 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.
January 4th, 2019 at 1:14:41 PM
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Based on the title of this thread, I thought the answer might be 8675309
January 4th, 2019 at 1:35:15 PM
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Quote: gamerfreakBased 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
If the House lost every hand, they wouldn't deal the game.
January 4th, 2019 at 1:47:52 PM
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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.Quote: Wizardln(-1) = 0
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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January 4th, 2019 at 1:54:14 PM
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Quote: gamerfreakBased on the title of this thread, I thought the answer might be 8675309
It almost was
January 4th, 2019 at 1:55:51 PM
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Quote: RSYou 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)
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 4th, 2019 at 2:06:28 PM
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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.Quote: WizardSure 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)
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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January 4th, 2019 at 2:31:00 PM
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Sadly, I'd understand this thread equally well if the whole thing was in Swahili.
The older I get, the better I recall things that never happened
January 4th, 2019 at 2:33:28 PM
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Or x=$200Quote: gamerfreakBased on the title of this thread, I thought the answer might be 8675309
I am a robot.
January 4th, 2019 at 3:25:32 PM
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Quote: WizardSure 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
January 4th, 2019 at 3:47:57 PM
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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)?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 4th, 2019 at 4:23:40 PM
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Quote: WizardNew 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.
January 4th, 2019 at 5:02:32 PM
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Kazi yako ya nyumbani ni kuthibitisha formula ya Euler: e ^ (bi) = cos (b) + i * sin (b)Quote: billryanSadly, I'd understand this thread equally well if the whole thing was in Swahili.
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January 4th, 2019 at 5:07:49 PM
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Quote: ThatDonGuyEr, 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)
