Let's Solve An Exponential Equation with Golden Flavor

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[@manwork6545]

easy, peasy!

[@nasrullahhusnan2289]

Let y=x²
The given equation becomes
(2^y)+(4^y)=8^y
--> (2^y)+(2^y)²=(2^y)³
Divide 2^y≠0 then put all terms in one side:
(2^y)²-(2^y)-1=0
is a equation for golden ratio ß, say
2^y = ß where ß=½(1+sqrt(5)]
y=²log(ß) where the log is based 2
x=±sqrt[²log(ß)]
The other root, -1/ß doesn't apply as 2^y can't be negative

[@SidneiMV]

2^x² = u (u > 0)
u² - u - 1 = 0
u = (1 + √5)/2
2^x² = (1 + √5)/2
x²ln2 = ln( 1 + √5) - ln2
x² = (1/ln2)ln(1 + √5) - 1
x = ± √[ (1/ln2)ln(1 + √5) - 1 ]

[@dalekloss4682]

you are right he forgot the ln2 or 1 in his solution.
in your last equation your forgot to change the ln2 to 1

[@p12psicop]

What are the answers in binary considering that you were using base 2?

[@rob876]

u = 2^(x^2)
u + u^2 = u^3
1 + u = u^2
u = (1 + √5)/2
x^2 = ln[(1 + √5)/2] / ln(2)
x = ±√{ln[(1 + √5)/2] / ln(2)}

[@DonEnsley]

x ∈ { -√[ln[½(1+√5)]/ln 2],
√[ln[½(1+√5)]/ln 2],
-√{ln[½(√5-1)]+iπ(1+2k)}/ln 2},
√{ln[½(√5-1)]+iπ(1+2k)}/ln 2}, ( k ∈ ℤ ) }