## Poll

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**6 members have voted**

(x

^{2}- 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?

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

For real numbers a and b, a

^{b}= 1 if:

(1) a = 1 -> x

^{2}- 15x + 55 = 1 -> (x - 6)(x - 9) = 1 -> x = 6, 9

(2) b = 0 -> x

^{2}- 9x + 20 = 0 -> (x - 4)(x - 5) = 0 -> x = 4,5

(3) a = -1 and b is even

a = -1 -> x

^{2}- 15x + 55 = -1 -> (x - 7)(x - 8) = 0 -> x = 7,8

x = 7 -> b = 7

^{2}- 9 * 7 + 20 = 6, which is even, so x = 7 is valid

x = 8 -> b = 8

^{2}- 9 * 8 + 20 = 14, which is even, so x = 8 is valid

Quote:ThatDonGuy

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

For real numbers a and b, a^{b}= 1 if:

(1) a = 1 -> x^{2}- 15x + 55 = 1 -> (x - 6)(x - 9) = 1 -> x = 6, 9

(2) b = 0 -> x^{2}- 9x + 20 = 0 -> (x - 4)(x - 5) = 0 -> x = 4,5

(3) a = -1 and b is even

a = -1 -> x^{2}- 15x + 55 = -1 -> (x - 7)(x - 8) = 0 -> x = 7,8

x = 7 -> b = 7^{2}- 9 * 7 + 20 = 6, which is even, so x = 7 is valid

x = 8 -> b = 8^{2}- 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

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.

Quote:WizardQuote:ThatDonGuy

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

For real numbers a and b, a^{b}= 1 if:

(1) a = 1 -> x^{2}- 15x + 55 = 1 -> (x - 6)(x - 9) = 1 -> x = 6, 9

(2) b = 0 -> x^{2}- 9x + 20 = 0 -> (x - 4)(x - 5) = 0 -> x = 4,5

(3) a = -1 and b is even

a = -1 -> x^{2}- 15x + 55 = -1 -> (x - 7)(x - 8) = 0 -> x = 7,8

x = 7 -> b = 7^{2}- 9 * 7 + 20 = 6, which is even, so x = 7 is valid

x = 8 -> b = 8^{2}- 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.

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

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

Quote:gamerfreakBased on the title of this thread, I thought the answer might be 8675309

It almost was

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)

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.

Or x=$200Quote:gamerfreakBased on the title of this thread, I thought the answer might be 8675309

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 e

^{x}with respect to x is e

^{x}

By Taylor's theorem (with respect to x = 0), e

^{bi}= 1 + bi + (bi)

^{2}/ 2! + (bi)

^{3}/ 3! + ...

= (1 + b

^{2}i

^{2}/ 2! + b

^{4}i

^{4}/ 4! + b

^{6}i

^{6}/ 6!+ ...)

+ (bi / 1! + b

^{3}i

^{3}/ 3! + b

^{5}i

^{5}/ 5! + b

^{7}i

^{7}/ 7! + ...)

= (1 - b

^{2}/ 2! + b

^{4}/ 4! - b

^{6}/ 6! + ...) + i (b - b

^{3}/ 3! + b

^{5}/ 5! - b

^{7}/ 7! + ...)

Again by Taylor's theorem, cos b (with respect to 0) = 1 - b

^{2}/ 2! + b

^{4}/ 4! - b

^{6}/ 6! + ...,

and sin b (with respect to 0) = b - b

^{3}/ 3! + b

^{5}/ 5! - b

^{7}/ 7! + ...),

which means cos b + i sin b = (1 - b

^{2}/ 2! + b

^{4}/ 4! - b

^{6}/ 6! + ...) + i (b - b

^{3}/ 3! + b

^{5}/ 5! - b

^{7}/ 7! + ...) = e

^{bi}

Quote:ThatDonGuy

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

The nth derivative of e^{x}with respect to x is e^{x}

By Taylor's theorem (with respect to x = 0), e^{bi}= 1 + bi + (bi)^{2}/ 2! + (bi)^{3}/ 3! + ...

= (1 + b^{2}i^{2}/ 2! + b^{4}i^{4}/ 4! + b^{6}i^{6}/ 6!+ ...)

+ (bi / 1! + b^{3}i^{3}/ 3! + b^{5}i^{5}/ 5! + b^{7}i^{7}/ 7! + ...)

= (1 - b^{2}/ 2! + b^{4}/ 4! - b^{6}/ 6! + ...) + i (b - b^{3}/ 3! + b^{5}/ 5! - b^{7}/ 7! + ...)

Again by Taylor's theorem, cos b (with respect to 0) = 1 - b^{2}/ 2! + b^{4}/ 4! - b^{6}/ 6! + ...,

and sin b (with respect to 0) = b - b^{3}/ 3! + b^{5}/ 5! - b^{7}/ 7! + ...),

which means cos b + i sin b = (1 - b^{2}/ 2! + b^{4}/ 4! - b^{6}/ 6! + ...) + i (b - b^{3}/ 3! + b^{5}/ 5! - b^{7}/ 7! + ...) = e^{bi}

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)?

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.

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.

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).