We'll substitute x^log2 x by t and we'll re-write the
equation in t:
t + 8*t^-1 =
6
We'll use the rule of negative
powers:
t +8/t = 6
We'll
multiply by t both sides:
t^2 + 8 =
6t
We'll move all terms to one
side:
t^2 - 6t + 8 = 0
Since
the sum is 6 and the product is 8, we'll conclude that the roots of the quadratic
are:
t1 = 2 and t2 = 4
x^log2
x = t1
x^log2 x = 2
We'll take
logarithms both sides:
log2 x^log2 x = log2
2
We'll apply the power rule of
logarithms:
log2 x^log2 x =
1
(log2 x)^2 - 1 = 0
We'll
re-write the differnce of squares:
(log2 x - 1)(log2 x + 1)
= 0
log2 x - 1 = 0 => log2 x=1 => x =
2^1
x = 2
log2 x + 1 = 0
=> log2 x = -1 => x = 2^-1
x =
1/2
x^log2 x = t2
x^log2 x =
2
log2 (x^log2 x) = log2
4
log2 (x^log2 x) = log2
2^2
log2 (x^log2 x) = 2log2
2
log2 (x^log2 x) = 2
log2
(x^log2 x) - 2 = 0
(log2 x)^2 - 2 =
0
(log2 x - sqrt2)(log2 x + sqrt2) =
0
log2 x - sqrt2 = 0
log2 x =
sqrt2 => x = 2^sqrt2
log2 x =- sqrt2 => x =
2^-sqrt2
x =
1/2^sqrt2
The solutions of the equation are:
{1/2 ; 2 ; 2^sqrt2 ; 1/2^sqrt2}.
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