## Poll

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3 members have voted

Wizard Joined: Oct 14, 2009
• Posts: 22046
March 15th, 2020 at 4:53:16 PM permalink
The question is simple:

9x + 12x = 16x

Find x.
It's not whether you win or lose; it's whether or not you had a good bet.
CrystalMath Joined: May 10, 2011
• Posts: 1827
March 15th, 2020 at 5:32:00 PM permalink
-infinity (actually just approaches equality). Otherwise, there's a solution somewhere between 1 and 2.
I heart Crystal Math.
DJTeddyBear Joined: Nov 2, 2009
• Posts: 10249
March 15th, 2020 at 5:38:01 PM permalink
I�ve never been able to wrap my head around non-whole exponents.

Non-whole roots give me even more headaches.

On the plus side, math problems like this will still be a PITA after the apocalypse. 🤪
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard Joined: Oct 14, 2009
• Posts: 22046
March 15th, 2020 at 6:07:51 PM permalink
I forgot to say the usual rules apply:

1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. Past winners who must chime in early, may PM me.
3. Beer to the first satisfactory answer and solution, subject to rule 2.

Also, I am indeed looking for a real number for credit.
It's not whether you win or lose; it's whether or not you had a good bet.
unJon Joined: Jul 1, 2018
• Posts: 1863
March 15th, 2020 at 6:54:09 PM permalink
Quote: Wizard

Also, I am indeed looking for a real number for credit.

As opposed to an imaginary or complex number?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard Joined: Oct 14, 2009
• Posts: 22046
March 15th, 2020 at 7:59:17 PM permalink
Quote: unJon

As opposed to an imaginary or complex number?

Yep, and infinity.
It's not whether you win or lose; it's whether or not you had a good bet.
CrystalMath Joined: May 10, 2011
• Posts: 1827
March 15th, 2020 at 8:17:16 PM permalink

9x+12x=16x
Divide everything by 9x
1+(12/9)x=(16/9)x
simplify
1+(4/3)x=(16/9)x
recognize that (4/3)x = (16/9)(x/2)
1+(16/9)(x/2)=(16/9)x

let n = (16/9)(x/2)
substitute n into above equation
1 + n = n2
n2 - n - 1 = 0
n = (1+-5(1/2))/2

(16/9)(x/2) = (1+-5(1/2))/2
ln[(16/9)(x/2)] = ln[(1+-5(1/2))/2]
(x/2)ln(16/9)=ln[(1+-5(1/2))/2]

x = 2ln[(1+-5(1/2))/2]/ln(16/9)

It is only valid for the positive root.
x = 2ln[(1+5(1/2))/2]/ln(16/9) ~=1.672721

Not that it makes any difference here, but (1+5(1/2))/2 is also the golden ratio.
I heart Crystal Math.
Wizard Joined: Oct 14, 2009
• Posts: 22046
March 15th, 2020 at 8:58:54 PM permalink
Quote: CrystalMath

9x+12x=16x
Divide everything by 9x
1+(12/9)x=(16/9)x
simplify
1+(4/3)x=(16/9)x
recognize that (4/3)x = (16/9)(x/2)
1+(16/9)(x/2)=(16/9)x

let n = (16/9)(x/2)
substitute n into above equation
1 + n = n2
n2 - n - 1 = 0
n = (1+-5(1/2))/2

(16/9)(x/2) = (1+-5(1/2))/2
ln[(16/9)(x/2)] = ln[(1+-5(1/2))/2]
(x/2)ln(16/9)=ln[(1+-5(1/2))/2]

x = 2ln[(1+-5(1/2))/2]/ln(16/9)

It is only valid for the positive root.
x = 2ln[(1+5(1/2))/2]/ln(16/9) ~=1.672721

Not that it makes any difference here, but (1+5(1/2))/2 is also the golden ratio.

I agree. However, weren't you previously a member of the beer club?
It's not whether you win or lose; it's whether or not you had a good bet.
CrystalMath Joined: May 10, 2011
• Posts: 1827
March 15th, 2020 at 9:10:01 PM permalink
Quote: Wizard

Quote: CrystalMath

9x+12x=16x
Divide everything by 9x
1+(12/9)x=(16/9)x
simplify
1+(4/3)x=(16/9)x
recognize that (4/3)x = (16/9)(x/2)
1+(16/9)(x/2)=(16/9)x

let n = (16/9)(x/2)
substitute n into above equation
1 + n = n2
n2 - n - 1 = 0
n = (1+-5(1/2))/2

(16/9)(x/2) = (1+-5(1/2))/2
ln[(16/9)(x/2)] = ln[(1+-5(1/2))/2]
(x/2)ln(16/9)=ln[(1+-5(1/2))/2]

x = 2ln[(1+-5(1/2))/2]/ln(16/9)

It is only valid for the positive root.
x = 2ln[(1+5(1/2))/2]/ln(16/9) ~=1.672721

Not that it makes any difference here, but (1+5(1/2))/2 is also the golden ratio.

I agree. However, weren't you previously a member of the beer club?

I�m almost certain I haven�t participated since these were a contest. Thanks for doing these, since it�s the math that attracted me to this site in the first place.
I heart Crystal Math.
Wizard Joined: Oct 14, 2009