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A Beautiful Exponential Equation from Romania
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- Опубликовано: 10 авг 2022
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Good problem! When you see everything on the left is a factor of 2 or 3 that tells you to substitute.
Thanks so much. I think your voice is better than yesterday.
Really a good Problems with full solution, Keep up
Pongo t=(3/2)^x,ottengo una equazione in t, 6t^2-13t+6=0,cioe t=2/3,t=3/2,che quindi danno x=1,x=-1
Concordo
Super awesome to substitute the exponential
I checked the general case and found that for any pair of bases such that (a^3x-b^3x)/((a^2x)b^x-a^x(b^2x)), if x=1 is a solution then x=-1 is also a solution.
This follows from:
(a^3x-b^3x)/((a^2x)b^x-a^x(b^2x))=(a^3-b^3)/((a^2)b-a(b^2))
Factoring as a difference of cubes and all that on both sides you get
(a^2x+(ab)^x+b^2x)/(ab)^x=(a^2+(ab)+b^2)/(ab)
Substitute in x=-1 and simplify to get b/a+1+a/b, which becomes (a^2+(ab)+b^2)/(ab) when you put it under a common denominator.
By observation (totally real and legot method):
x=1
Imagine taking 9 minutes to count to 1
@@jeremymorain you do realise that i have a life outside youtube and that when i got on my phone the video was already 9 minutes? Yeah i guessed it, you don't.
Genial gracias
By a look, we can definitely say that 1 is a solution.
lol yes
There is another method, which takes advantages of function knowledge.
Let f(x) = (27^x - 8^x) / (18^x - 12^x)
By plugging -x in, we will have f(-x) = [27^(-x) - 8^(-x)] / [18^(-x) - 12^(-x)] and this looks kind of complicated, so let's simplify this a bit.
f(-x) = [27^(-x) - 8^(-x)] / [18^(-x) - 12^(-x)]
= (1/27^x - 1/8^x) / (1/18^x - 1/12^x)
= [(8^x - 27^x) / 216^x] / [(12^x - 18^x) / 216^x]
= (27^x - 8^x) / (18^x - 12^x)
Since f(x) = f(-x), f(x) is an even function, meaning if x is a solution, -x is also a solution. Now we need to prove f(x) is strictly increasing for x > 0 by using differentiation (a.k.a the longest part of the method).
f'(x) = {(18^x - 12^x)[27^x.log(27) - 8^x.log(8)] - (27^x - 8^x)[18^x.log(18) - 12^x.log(12)]} / (18^x - 12^x)^2
Let the numerator be g(x) and we'll factorize this madness.
g(x) = 6^x(3^x - 2^x)[27^x.log(27) - 8^x.log(8)] - 6^x(3^x - 2^x)(4^x + 6^x + 9^x)[3^x.log(18) - 2^x.log(12)]
= 6^x(3^x - 2^x){27^x.log(27) - 8^x.log(8) - (4^x + 6^x + 9^x)[3^x.log(18) - 2^x.log(12)]}
= 6^x(3^x - 2^x).h(x)
Now we further simplify h(x) by expanding them and putting them back (I hope I use this phrase correctly).
h(x) = 27^x.log(27) - 8^x.log(8) - 12^x.log(18) - 18^x.log(18) - 27^x.log(18) + 8^x.log(12) + 12^x.log(12) + 18^x.log(12)
= 8^x[log(12) - log(8)] - 12^x[log(18) - log(12)] - 18^x[log(18) - log(12)] + 27^x[log(27) - log(18)]
= 8^x.log(12/8) - 12^x.log(18/12) - 18^x.log(18/12) + 27^x.log(27/18)
= 8^x.log(3/2) - 12^x.log(3/2) - 18^x.log(3/2) + 27^x.log(3/2)
= [log(3/2)](8^x - 12^x - 18^x + 27^x) = log(3/2)(2^x - 3^x)^2(2^x + 3^x)
Therefore, f'(x) = [6^x.log(3/2)(2^x + 3^x)(3^x - 2^x)^3] / (18^x - 12^x)^2
Because x > 0, 3^x - 2^x > 0 and (18^x - 12^x)^2 ≠ 0. Therefore, f'(x) > 0, which means f(x) is even and for x > 0, it is strictly increasing. Now we just need to guess and conclude to the other solution.
Seeing that f(1) = f(-1) = 19/6, that tells us that x = ±1.
No, even function doesn't mean that this equation has exactly 2 opposite solutions !
x=1 is an obvious solution then x=-1 is also a solution now you must prove that is unique. You must prove that for x >0 then the function is strictly increasing (or decreasing) or injective
@@WahranRai oh, right, but at least that part is not too hard to do after all 😅 i'll include it in later
An interesting method.
Excelente 👍👍👍
Nicely done! I'm a couple of days behind in the videos. I hope you are feeling better soon, my friend.
Thank you!!! 🥰🧡
يعجبني هذا نوع من اسئله شكرا عله جهودكم٨
Спасибо за видео!!!
Np
very nice question
About the graph of y = f(x) = (27^x − 8^x)/(18^x − 12^x) and its symmetry...
First, note that the function is not defined at 0.
When 0 is excluded and we divide numerator and denominator by (3^x − 2^x), we get y = (9^x + 6^x + 4^x)/6^x, i.e. y = (3/2)^x + 1 + (2/3)^x. Then it is not surprising that the function is symmetric around 0, because replacing x with −x is equivalent to swapping (2/3) and (3/2).
There is more : f(x)-1 = 1.5^x + 1.5^-x is very similar to cosh(x) = e^x + e^-x !
Oops! I mean cosh(x] = (e^x + e^-x)/2. By the way, our function f(x) is exactly the same as 1+2.cosh(x ln(3/2)) for all x except 0.
👍
Your sound looks like a different..seems you got flow, hope you recover soon ان شاء الله
Yeah thanks 🥰
Awesome ♥️
Sir, your voice sounds a bit different 😅
Thank you! yes 🤒
Hi~!
@@SyberMath oops.. Get well soon 💖
@@SuperYoonHo Hi sir! how's going everything?
@@jimmykitty Hi everything is find EXCEPT we didn't have the internet for 2 days again!!! haha @SyberMath you know what i mean do you?
it is a mouse that nibbled on our wire
sorry for the late reply...
How is it going there? Here is fine now:)
When you get
6a^2-13ab+6b^2 = 0,
the factorization is not at all obvious.
So instead, since a, b are both not zero, I divided both sides by ab and obtained:
6(a/b)-13+6(b/a) = 0
Substituting w=a/b we get :
6w-13+6(1/w) = 0
6w^2-13w+6 = 0
Solving this quadratic equation we obtain :
w = 3/2 or w = 2/3
I would rather divide both sides by b² or a² instead, it will instantly create a quadratic equation.
@@gdtargetvn2418 I followed this approach as well
@@gdtargetvn2418 I divided by ab, then multiplied by a/b, which is equivalent to dividing by b^2, so you're right, I might have done it in a single step dividing by b^2. Of course dividing by a^2 is also possible.
Cool
thnku
Np
Hello:D
@@SuperYoonHo hello
@@pranavamali05 good afternoon this has been a busy day with my animals im tired...
@@SuperYoonHo have a good sleep tonight
how does 3^x/2^x =3/2
admission test for 11 year boys to math classes in USSR.
Have you played Club Penguin or Club Penguin Rewritten before?
x=1 and x=-1 answer
x=±1
X=1
Ans x= 1 (no need to calculate/)
Not the only solution to the equation. Many of these problems are like that -- you can find one or two solutions easily, but the hard part is finding them all (or proving that there are no more).
👍👍👍👏👏👏👌👌👌
x = 1. Are you Baldi?
I find no real solution :(
Are you sure?
@@SyberMath I'd find x = 0 and a quadralic which have no real solution