A Logarithmic System with a Golden Flavor

Рет қаралды 18,256

Күн бұрын

Пікірлер: 40

@Skank_and_Gutterboy 2 жыл бұрын

One thing you can look at with Desmos, too: plot the two curves y=x^ln(x) and y=(x-1)^ln(x-1). The two curves intersect at x=1.618... which corresponds to (sqrt(5)+1)/2. That x-coordinate corresponds to a. To get b, b=a-1.

@rajeshbuya 2 жыл бұрын

Very nice 👍 Golden Problem.. very interesting to see how 'ln' and 'phi' are connected. Now, that's MatheMagic!!!

@Skank_and_Gutterboy 2 жыл бұрын

6:27 Just for grins, I took b=(-1-sqrt(5))/2 and calculated it through the system, just kind of ignoring complex results and running with it anyway. It doesn't satisfy the system, it can't. If you force it, you'll just wind up eliminating it later as a solution anyway.

@ogy2014 Жыл бұрын

but it does

@pirvumihaiteodor1766 2 жыл бұрын

The graph looks like parabolica in Monza Circuit F1

@miikey_lol 2 жыл бұрын

Yeah it kind of does look like it.

@mcbeaulieu 2 жыл бұрын

Mirrored but still true 😅

@pavelkotsev1542 2 жыл бұрын

In two races time we're gonna have the sin function named after Senna... just sayin' :)

@tontonbeber4555 2 жыл бұрын

No need for a paper and pencil ... easy enough to solve mentally ... By taking ln of (1) => ln(a^lna) = ln(b^lnb) => (lna)^2 = (lnb)^2 Obviously we can't have lna = lnb (because then a=b and (2) is impossible) => lna = - lnb => a = 1/b And then we have the usual golden a - 1/a = 1 => a2 - a -1 = 0 => a = (1+V5)/2 and (2) gives b = a-1 = (-1+V5)/2 (of course the negative solution a = (1-V5)/2 is rejected because of ln)

@Skank_and_Gutterboy 2 жыл бұрын

ln(a) = -ln(b), that's where I got tripped up. After getting that concept, it all fell into place. This is a great problem!

@subhashkumarsinha8252 Жыл бұрын

Excellent problem and solved beautifully

@yoav613 2 жыл бұрын

Very nice and creative problems latly,great job!!😃💯💥

@SyberMath 2 жыл бұрын

Thank you very much! 🤩🤓

@broytingaravsol 2 жыл бұрын

a=(√5+1)/2, b=(√5-1)/2

@mathswan1607 2 жыл бұрын

Solving a-b=1 and ab=1 gives a=(1+root(5))/2 and b=(-1+root(5))/2

@davidseed2939 7 күн бұрын

. lna.lna=lnb.lnb this gives lna=+-lnb clearly if a=b+1 then lna= lnb is impossible. hence lna =-lnb so lna +lnb=0 or ln(ab)=0 or ab=1 so a=1+1/a or a^2-a-1=0 or a= (-1+- sqrt(5))/2

@haliltokmakmatgeo 2 жыл бұрын

a=(√5+1)/2 b=(√5-1)/2🙂🙂

@FenetreSurLeMonde 2 жыл бұрын

a^ln(a)=b^ln(b) f(x)=x^ln(x)=[e^(ln(x))]^ln(x)=e^(ln(x))² f'(x)=(2*ln(x)/x)f(x) So if 0

@shhgs 2 жыл бұрын

a = e ^ {ln a} => a ^ {ln a} = e ^ {{ln a} ^ 2} => ln a = - ln b

@bars-matik4818 2 жыл бұрын

f(x^3)=3f(x) can you do it?

@morteza3268 2 жыл бұрын

F(x)=ln(x) 🙂

@vatsal941 2 жыл бұрын

Not only lnx .. log with any base will go true for this functional equation

@daddykhalil909 2 жыл бұрын

Excellent

@tbg-brawlstars 2 жыл бұрын

a = (√5 + 1)/2 , b = (√5 - 1)/2 Also, I am first

@davidseed2939 2 жыл бұрын

your soln=x^lnx I see now that (1,1) is a minimum of the curve and f(x)=f(1/x). so i offer this alternative function. f(x)=1/2(x^-2 + x^2) it looks very similar. if you had said (e,e) is on the curve, i might have got it.

@SyberMath 2 жыл бұрын

It does look very similar!!! Wow!

@davidseed2939 2 жыл бұрын

but there is also a value of a in the following which does pass through (e,e) f(x)=1/2(x^-a+x^a) solve for a … e=(1/2)(e^-a+e^a)=cosh(a) a=arcosh(e)~=1.6574