A Golden Equation #algebra

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[robertlunderwood]

Instead of taking ln, do log base 2. It'll yield a cleaner solution.

[graf_paper]

Wow, I solved this before watching and was pretty surprised at the difference in method.

I divided both sides by (2√2)ˣ and rearranged to get

√(2⁻ˣ) + (2⁻ˣ) = 1

I then subbed in u = (2⁻ˣ)

√u + u = 1
√u = 1 - u
u = 1 -2u + u²
u² - 3u + 1 = 0

u = (3 + √5)/2

(2⁻ˣ) = (3 + √5)/2

1/2ˣ = (3 + √5)/2

Cross multiplying we get

2^(x+1) = 3 + √5

Log₂(3 + √5) = x + 1

Log₂(3 + √5) - 1 = x

1.39 ≈ x

I guess I just made a slightly different substitution and used a log with a different base. Thanks for the problem! Great video.

[prollysine]

let u=(V2)^x, u+u^2=u^3, u^3-u^2-u=0, u(u^2-u-1)=0, u=0, (1+V5)/2, (1-V5)/2, (V2)^x=(1+V5)/2,
/ u=0 & u=(1-V5)/2 not a solu /,
x=log((1+V5)/2)/log(V2), test, (V2)^x+2^x=~ 4 .23607, (2*V2)^x=~ 4 .23607, OK,