Solve for x x^(lnx/ln2)=6-8/x^(lnx/ln2)

We'll substitute x^(ln x/ln 2) by t and we'll re-write the
equation in t:

t  = 6 -
8/t

We'll multiply by t both
sides:

t^2 = 6t - 8

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^(ln x/ln 2) = t1

x^(ln
x/ln 2) = 2

But ln x/ln 2 = log 2
x

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^(ln x/ln 2) = t2

x^(ln
x/ln 2) = 4

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

All values of x that verify the
equation are: {1/2 ; 2 ; 2^sqrt2 ; 2^-sqrt2}.