Question
Find the value of 12 to the power of (-1+sqrt(5))/2
\Big( \dfrac{-1+\sqrt{5} }{2} \Big) ^{12}
Answer 1
Let
a = \dfrac{-1+\sqrt{5} }{2}
(1)
Rearrange the equation
2a+1 = \sqrt{5}
Square both sides
4a^2+4a+1=5
Then,
a^2=1-a
(2)
Square both sides
a^4 = a^2-2a+1
(3)
Substituting (2) to (3) gives
a^4= 2-3a
(4)
Multiplying (2) by (4) gives
a^6 = (1-a)(2-3a)
Expand the right hand side
a^6 = 2-5a+3a^2
(5)
Substituting (2) to (5) gives
a^6 = 2-5a+3-3a = 5-8a
(6)
Squaring both sides of equation (6) gives
a^{12} = 25-80a+64a^2
(7)
Substitute (2) to (7)
a^{12} = 25-80a+64(1-a) = 89-144a
(8)
Substitute the value of a to (8), we get final result
a^{12} = 89 - 144\cdot \dfrac{-1+\sqrt{5} }{2} = 161 - 72\sqrt{5}
that is, 12 to the power of (-1+sqrt(5))/2 is euqal to 161 - 72\sqrt{5}
Answer 2
Let
a = \dfrac{-1+\sqrt{5} }{2}
(1)
Then,
\dfrac{1}{a} = \dfrac{1}{ \dfrac{-1+\sqrt{5} }{2}} = \dfrac{1+\sqrt{5} }{2}
(2)
Addition of a and \dfrac{1}{a} gives a Nike equation
a+\dfrac{1}{a} = \sqrt{5}
(3)
Square both sides and move the constant term to the right
a^2+\dfrac{1}{a^2} = 3
(4)
Square both sides and move the constant term to the right
a^4+\dfrac{1}{a^4} =7
(5)
Multiplying (2) by (3) gives
a^6+a^2+\dfrac{1}{a^2} +\dfrac{1}{a^6} = 21
(6)
Substitution (2) to (4) gives
a^6+\dfrac{1}{a^6} = 18
(7)
Squaring both sides of (5) and move the constant term to the right
a^{12}+\dfrac{1}{a^{12}} = 18^2-2=322
(8)
Let
m =a^{12}
(9)
Then a quadratic equation is obtained
m^2-322m+1 = 0
Using the root formula for a quadratic equation
m = \dfrac{322\pm\sqrt{322^2-4} }{2}
Since \dfrac{-1+\sqrt{5} }{2} <1, cancel the root larger than 1. Then we obtained the final result
m = 161-72\sqrt{5}