(6-2sqrt5)^x - 5(sqrt5-1)^x + 6 = 0

Let us assume that :

y= (sqrt5 - 1)^x

Square both sides:

==> y^2 = (sqrt5 - 1)^2x = (6- 2sqrt5)^x

==> y^2 - 5y + 6 = 0

Now factor:

( y -3)(y-2) = 0

y1= 3 ==> 3= (sqrt5 - 1)^x ==> x1 = log (sqrt5-1)/log 3

y2= 2 ==> 2= (sqrt5 - 1)^x ==> x2= log (sqrt5 -1)/log 2

This is an exponential equation and we'll solve it using the substitution technique.

We notice that if we'll raise to square (sqrt5-1 )^2, we'll get:

(sqrt5-1 )^2 = 5 - 2sqrt5 + 1 = 6 - 2sqrt5

So, we'll rewrite the given equation:

(sqrt5-1 )^2x - 5(sqrt5-1 )^x + 6 = 0

We'll note (sqrt5-1 )^x by t:

t^2 - 5t + 6 = 0

We'll apply the quadratic formula:

t1 = [5 + sqrt(25-24)]/2

t1 = (5+1)/2

t1 = 3

t2 = (5-1)/2

t2 = 2

(sqrt5-1 )^x = 3

We'll take logarithms:

lg (sqrt5-1 )^x = lg 3

We'll use the power property of logarithms:

x *lg (sqrt5-1 ) = lg 3

We'll divide by lg (sqrt5-1 ):

**x1 = lg 3 / lg (sqrt5-1 )**

(sqrt5-1 )^x = 2

We'll take logarithms:

lg (sqrt5-1 )^x = lg 2

We'll use the power property of logarithms:

x *lg (sqrt5-1 ) = lg 2

**x2 = lg 2 / lg (sqrt5-1 )**

(6-2sqrtx) -5(sqrt5 -1) +6 = 0

To solve for x

We know that (sqrt5-1)^2 = 5 -2sqrt5 +1^2 = (6-2sqrtx).

So the given equation could be wroteen as:

( y^2)^x-5y^x+6 = 0, where y=sqrt5 -1.

(y^x)^2 - 5(y^x) +6 = 0

(y^x -2)(y^2-3) = 0

y^x = 2 Or y^x =3.

(sqrt5 -1)^x =2

x log(sqrt5-1) = log2

x = log2/log(sqrt5-1).

Similarly (sqrt5-1)^x = 3 gives: x = log3/log(sqrt5-1)