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
(The entire section contains 112 words.)
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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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