How to solve this unbelievable exponential equation

We will solve a (tricky?) triple exponential equation e^x+e^(2x)=e^(3x). We need to know the properties of exponential functions, usage of the quadratic formula, and usage of the logarithm. We will see the golden ratio at the end!
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#math #algebra #mathbasics

[familyguytimes]

this guy makes me love math even more each time I watch him!

[the_magisterate]

when i first saw this question and tried to solve it, I divided by e^2x on both sides. Interestingly, after rearranging the variable terms, we can get the equation for sinh. idk if it's cheating but I just took the arcsinh(1/2) to get the same answer.

I think it's pretty cool that seemingly unrelated mathematical concepts like hyperbolic trig and golden ratio can unexpectedly unify

[supramayro]

at the end of the video he went from 😁 to 😒

[marcushendriksen]

Let e^x = y. Then the equation becomes y + y^2 = y^3, which is the same as y^3 - y^2 - y = 0. Factor out a y: y(y^2 - y - 1) = 0. Therefore y = 0 or y^2 - y - 1 = 0. The second equation leads to y being (1+sqrt5)/2 or (1-sqrt5)/2.

Since e^x = y, y cannot be 0 or negative (since the real logarithm function is undefined for those), so x must be ln(1+sqrt5) - ln2

[andrewmccauley]

x^n = x^(n-1) + x^(n-2) is a fundamental property of golden ratio and related to fibonacci and lucas series, which I love. So I was on to this one pretty quickly, but still cool to see.

[januszjasniowski]

Dziękuję za wykład zw. z liczbą e.
Wspaniały matematyk i wykładowca. Jeszcze raz serdecznie dziękuję. Oglądam pana wykłady z matematyki od kilku lat.
Mam bardzo duży zbiór zadań z matematyki średniej i wyższej oraz z olimpiad na joutube.

[ScorpioHR]

I just wanted to ask "what's with PokéBall" on the newest video and now I see it's missing. So I have a following question: What happened to the PokéBall?

[LearnWithFardin]

x=sinh-¹(1/2)
My approach:

e^(x) + e^(2x) =e^(3x)

1+e^(x)=e^(2x)

(e^x)[e^x -1]=1

e^(x) - e^-(x) =1

2sinh(x)=1

Therefore,

x=sinh-¹(1/2)


[I know, this is an algebra channel but anyway]

[maths]

The complex solutions are: ln((-1+√5)/2)+i*(2kπ+π) and ln((1+√5)/2)+2ikπ where k∈Z. Thanks to KasabianFan44 for the addition!

[ciberiada]

Wow! An elegant problem and solution!
I prefer avoiding the fraction and write it:
ln(1+√5) – ln(2)
As √5 ≈ 2.24 and √5 + 1 ≈ 3.24, so ln(1+√5) > 1 and ln(2) < 1
let's say ln(1+√5) = 1 + δ and
ln(2) = 1 – ε, where δ and ε are < 0.5
So 1 + δ – 1 + ε = δ + ε < 1

[holyshit]

ln(lim(F_{n+1}/F_{n}, n=infinity))
where F_{n} is Fibonacci sequence

[jessetrevena]

Great video! Very satisfying solution.

[sebicastian]

You are awesome, i thought i will resolve by my self, but now i realise that much terms in math that i need to know for get your level in math.


Great video and im sorry if my english is not good for write :)

[Robert_H.]

log(1 + e^x) = log(e^(2x))^= 2x

log(1 + e^x) = Log(2) + x/2 + x^2/8 + ...

x^2/8 - 3x/2 + Log(2) = 0

x = 6 - 2 sqrt(9 - log(4)) = 0.48141118...

Comparision to exact solution:

x = log(phi) = 0.4812118...

So in case you haven't seen the substitution u = e^x, you do come up with a very accurate solution using a Taylor approximation.

[quevineuxcrougniard]

Ce chinois est tout a fait compréhensible grâce à son tableau en dépit de son baragouin imbittable ! Alors je m'abonne et je le remercie.

[dinuwarabinudithdesilva]

That is the fabulous element, the gold(I mean the golden ratio)

[rageprod]

If we take it to the complex game (which we should, because it's fun), we have:

e^x = φ, therefore x = lnφ (as shown)

but also, e^x = (1-sqrt5)/2 = - 1/φ (demonstration left as exercise :P), therefore

x = ln(- 1/φ) = ln(-1) + ln(1/φ) = ln(-1) - lnφ (by the property of logarithms of exponents)

but what the heck is ln(-1)?

recall e^iπ(2k+1) = -1, for k in Z; we take the natural logarithm of both sides:

ln(-1) = iπ(2k+1)

therefore: x = iπ(2k+1) - lnφ, for k in Z

So we have one real solution and infinitely many complex solutions, because of the complex logarithm's periodic property! Very nice.


Edit: sign mistake, oops

As corrected by Michael Vogel, e^iπ2k = 1, so ln1 = iπ2k for k in Z.

lnφ = ln1 + lnφ, so iπ2k + lnφ for k in Z are also complex solutions!

[amankumarsharma]

Amazing solving exponential eq through Quadratic eq properties.
Maths is so beautiful 🥰

[undsamuel]

I have a question (out of real solutions). If we replace e^x = u, then we would have u+u^2 = u^3, does that mean that the first equation has 3 solutions? 1 real solution and 2 complex solutions? Because when I try to solve all the 3 solutions by factoring I get an impossible solution: e^x = 0, which is not possible. So does that mean that, even if you have an equation of grade 3, maybe you don't get 3 solutions even in complex numbers?

[PragmaticAntithesis]

Here's my solution. I'll see if I'm correct after watching the video.
Let y=e^x
Let g be the golden ratio (1.618...)

y+y^2=y^3
This has three solutions: y=0, y=g, y=-1/g
However, because y is an exponential, y>0. Thus, y=g.
Solving for x: e^x=g, so x=ln(g), which is about 1/2.