Solving e³ˣ – e²ˣ = eˣ | The Golden Ratio

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[SyberMath]

Sorry, I totally forgot to share it. Here is the graph:

[wtspman]

You promised us a graph!

Being serious: thanks for digging into the complex solutions for these problems. Stopping at the Real solutions implies something: complex numbers aren’t legitimate? Audience isn’t sophisticated enough? Something! Digging into the complex solutions acknowledges the curiosity that at least some of your audience will have, and it helps them build a bigger toolkit of problem-solving skills.

Thanks!

[Psykolord]

Before watching:

e^(3x)-e^(2x) = (e^(2x))(e^x -1) = e^x.

Let u = e^x. then e^(2x) = u^2.

-> u^2(u-1) = u.

Divide both sides by u (we can do this because we know e^x >0 for all real x)

u(u-1) = 1 = u^2-u.

Add 1 to both sides:

u^2 - u -1 = 0.

Now, we use the quadratic formula, with a=1, b=-1, = -1

(1±√5)/2 will be our values for u. However, since u = e^x, and e is positive, that means e^2x cannot be negative. Thus, (1-√5)/2 is not a potential solution.

Thus e^x = (1+√5)/2, take ln of both sides, and:

x = ln ((1+√5)/2) is our only *real number* solution.

A couple of my initial steps were unnecessary, looking back on it, but it works either way.

[quark]

Before using the substitution u=e^x, you can simplify a bit your equation. As e^x is never 0, you can divide both side of the beginning equation by e^x. So after substitution with u, the equation becomes u²-u=1, so you will never have u=0 as a solution. Also, when asking a solution for an equation, you need before to say in which set you search the solution. In the real set? In the complex set? In the set of matrix of real or complex entries? Because yes, e^M, where M is a matrix, has a meaning :). We can consider that if you don't say in which set you search the solutions of an equation, you search in ℝ :).

[groscolisdery]

I was going to try it when I unfortunately saw the beginning of your demonstration.

Third degree equation.

Yes, I was going to start with u=exp(x).

u=0 Is an obvious solution
that is for x-> - infinite

else we get u^2-u=1
(u-0.5)^2= 1+(0.5)^2

u= 0.5 +- ✓1.25

u=1, 6180339887498

u=-0, 6180339887498

for the positive root,
x= ln(1, 6180339887498)


for the negative root, x is a complex number

[writerightmathnation]

Around 2:03, you have a contradiction, because you wrote u_1=0 and then u_1=\frac{1+\sqrt5}2. This implies that 0>0, which is a contradiction. I'm guessing it's just an unintentional error, and you meant to name the three roots of the cubic equation three different names.

[lawrencejelsma]

Even in complex numbers (e^(0) + e^(-0))/2 is not equal to 0 for when (e^(0) - e^(0))/(2i) would have been 0. With Sin(0) or Sinh(0) = 0 there is no Cos(0) nor Cosh(0) = 0 there is no Cos = 0 but at an x value of π/2 + nπ real 0 but no imaginary 0 follows except an i and -i of magnitude 1.

[HoSza]

why stopping at complex? there are lots of other kinds of numbers also

[michaelaldam]

If you are considering all complex solutions, in addition to the golden ratio (\phi), you should have:

\phi + 2πni,

where n is an integer.

[MisterPenguin]

I DEMAND GRAPHS AT THE END OF EVERY VIDEO*!

* - I have no power here, I just like when you scroll down and Desmos is there... like a mathematical golden retriever.

[PKMath]

Seems like you promised a graph of it? Haha nice work Syber!

[darksamoth]

Why is there two solutions when the equation is cubic ?

[scottleung]

The real part is easy, but the complex part is a little far-fetched.

[lorenzbroll]

LOL. 3x-2x = 1x. Try a bit harder maybe?