An Interesting Cubic Exponential Equation
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- Опубликовано: 17 июн 2024
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Here’s a nice math problem which I came up with: find all pairs of primes p and q such that p^2 + q^3 is a perfect cube.
You can use that as a video idea if you want to
Thank you very much for the question❤. Very fascinating.
I don’t think syber will make a video on it tho, it’s too number theoretic. His videos are more algebra and calculus.
Thanks) Actually this equation only has a single solution, I find it beautiful
Nice!
L'equazione diventa una cubica...(log5)x^3+(log3)x^2+x-log30=0,con log base 2...le 3 soluzioni sono x=1,riduco l'equazione ad una quadratica, (x-1)(log5x^2+log15x+log30)=0...e 2 complesse
Lets go i got all 3 solutions
x = 1
Got x=1 in seconds but didn't work out the complex solutions 😂
Common base e, combine, cubic with logged coefficients = ln30. x=1 by inspection, divide by x-1 and quadratic formula.
I'm having a hard time seeing how the long division plays with the ln identities in my head, but I don't have the luxury of pen/paper right now.
...the main thing being that the four log constants have the property a+b+c=d
[2^(x - 1)][3^(x² - 1)][5^(x³ - 1)] = 1
(x - 1)ln2 + (x² - 1)ln3 + (x³ -1)ln5 = 0
x - 1 = 0 => *x = 1*
ln2 + (x + 1)ln3 + (x² + x + 1)ln5 = 0
(ln5)x² + (ln15)x + ln30 = 0
*x = [ - ln15 ± √(ln²15 - 4ln5ln30) ] / (2ln5)*
simplifying - an attempt
*x = [ - ln15 ± i√( 4ln5ln10 - ln²(5/3) ) ] / (2ln5)*
Just by looking at it.😅
On the last line, 2ln(5) must go inside of grouping symbols because of the Order of Operations.
@@robertveith6383 you are right. tks
The product of 3 positive increasing functions! What else do you need ?
Complex solutions
@@otaviogoncalvesdossantos862 which are obvious once x=1 is a root. Physical work.
Ln everything clean it up as a cubic in x and notice 1 is a solution so factor out x-1 and you get a quadratic with logs, the solutions are really messy and are complex so let's just ignore those and say 1 only solution
By comparing both sides you get x = 1, x^2 = 1 and x^3 = 1. So, the quadratic term gives 2 solutions x = 1 and x = -1... but the last one doesn't work.