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

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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

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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)

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