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

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### 2 Answers

We have to solve for x given that x^(ln x / ln 2) = 6 - 8/x^(ln x / ln 2).

x^(ln x / ln 2) = 6 - 8/x^(ln x / ln 2)

Let x^(ln x / ln 2) = y

=> y = 6 - 8/y

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

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

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

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

=> y = 2 and y = 4

As y = x^(ln x / ln 2)

=> y = ln(2) x

x^[ln(2) x] = 2

taking the ln(2) of both the sides

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

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

=> x = 2 and x = 1/2

x^[ln(2) x] = 2^2

taking the ln(2) of both the sides

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

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

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

**The required solution is x = 2, x = 1/2, x = 2^(sqrt 2) and x = 2^(-sqrt 2)**

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}.**