Let's Solve An Exponential Equation with Golden Flavor
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- Äas pĆidĂĄn 5. 07. 2024
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easy, peasy!
Let y=xÂČ
The given equation becomes
(2^y)+(4^y)=8^y
--> (2^y)+(2^y)ÂČ=(2^y)Âł
Divide 2^yâ 0 then put all terms in one side:
(2^y)ÂČ-(2^y)-1=0
is a equation for golden ratio Ă, say
2^y = Ă where Ă=Âœ(1+sqrt(5)]
y=ÂČlog(Ă) where the log is based 2
x=±sqrt[ÂČlog(Ă)]
The other root, -1/Ă doesn't apply as 2^y can't be negative
What are the answers in binary considering that you were using base 2?
2^xÂČ = u (u > 0)
uÂČ - u - 1 = 0
u = (1 + â5)/2
2^xÂČ = (1 + â5)/2
xÂČln2 = ln( 1 + â5) - ln2
xÂČ = (1/ln2)ln(1 + â5) - 1
x = ± â[ (1/ln2)ln(1 + â5) - 1 ]
you are right he forgot the ln2 or 1 in his solution.
in your last equation your forgot to change the ln2 to 1
x â { -â[ln[Âœ(1+â5)]/ln 2],
â[ln[Âœ(1+â5)]/ln 2],
-â{ln[Âœ(â5-1)]+iÏ(1+2k)}/ln 2},
â{ln[Âœ(â5-1)]+iÏ(1+2k)}/ln 2}, ( k â †) }
u = 2^(x^2)
u + u^2 = u^3
1 + u = u^2
u = (1 + â5)/2
x^2 = ln[(1 + â5)/2] / ln(2)
x = ±â{ln[(1 + â5)/2] / ln(2)}