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

[@p12psicop]

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

[@gelbkehlchen]

Solution: 2^(x²)+4^(x²) = 8^(x²) |/2^(x²)≠0 ⟹ 1+2^(x²) = 4^(x²) ⟹ 1+2^(x²) = 2^(2*x²) |with u = 2^(x²) ⟹ 1+u = u² |-u-1 ⟹ u²-u-1 = 0 |p-q-formula ⟹ u1/2 = 1/2±√(1/4+1) = 1/2±√(5/4) = (1±√5)/2 ⟹ u1 = (1+√5)/2 and u2 = (1-√5)/2 < 0 ⟹ 1st case: 2^(x1²) = u1 = (1+√5)/2 |lb() ⟹ x1² = lb[(1+√5)/2] |√() ⟹ x11/2 = ±√{lb[(1+√5)/2]} ≈ ±0,8332 x11 ≈ +0,8332 x12 ≈ -0,8332 2nd case: 2^(x2²) = u2 = (1-√5)/2 < 0 [that is not defined]

[@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

[@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)}