# x^log2 x + 8*x^-log2 x = 6. What is x?

### 2 Answers | Add Yours

We have to solve for x given that x^log2 x + 8*x^-log2 x = 6

x^log(2) x + 8*x^(- log(2) x) = 6

=> x^log(2) x + 8/x^(log(2) x) = 6

Let x^log(2) x = y

=> y + 8/y = 6

=> y^2 - 6x + 8 = 0

=> y^2 - 4y - 2y + 8 = 0

=> y(y - 4) - 2(y - 4) =0

=> ( y - 2)(y - 4) =0

We get y = 2 and y = 4

x^log(2) x = 2

=> log(2) x * log(2) x = 1

=> [log(2) x]^2 = 1

=> log(2) x = 1 and log(2) x = -1

=> x = 2 and x = 1/2

x^log(2) x = 4

=> [log (2) x]^2 = 2

=> log (2) x = sqrt 2 and - sqrt 2

=> x = 2^ sqrt 2 and 2^(- sqrt 2)

**The values of x are 2 , (1/2) , 2^ sqrt 2 and 2^(-sqrt 2)**

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}.**