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Wizard
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March 15th, 2020 at 4:53:16 PM permalink
The question is simple:

9x + 12x = 16x

Find x.
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CrystalMath
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March 15th, 2020 at 5:32:00 PM permalink
-infinity (actually just approaches equality). Otherwise, there's a solution somewhere between 1 and 2.
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DJTeddyBear
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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. 🤪
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard
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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.
  4. Please put answers and solutions in spoiler tags.


Also, I am indeed looking for a real number for credit.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
unJon
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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?
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Wizard
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March 15th, 2020 at 7:59:17 PM permalink
Quote: unJon

As opposed to an imaginary or complex number?



Yep, and infinity.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
CrystalMath
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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.
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Wizard
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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?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
CrystalMath
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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
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March 16th, 2020 at 12:44:57 PM permalink
Quote: CrystalMath

Thanks for doing these, since it’s the math that attracted me to this site in the first place.



You're welcome!
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ThatDonGuy
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March 16th, 2020 at 5:00:47 PM permalink

Let a = 3x and b = 4x
The original equation can be written as:
a2 + ab = b2
a2 + ab + b2 / 4 = b2 + b2 / 4
(a + b/2)2 = 5 / 4 * b2
a + b/2 = sqrt(5) / 2 * b
a = (sqrt(5) - 1) / 2 * b
Substitute back for a and b:
3x = (sqrt(5) - 1) / 2 * 4x
(3/4)x = (sqrt(5) - 1) / 2
x log (3/4) = log((sqrt(5) - 1) / 2)
x = log((sqrt(5) - 1) / 2) / log (3/4) = 1.672721

Note that the logarithms can be of any valid base, provided of course that they are all the same base

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