What is x if lg^2 x^3 - 6lg x + 1 = 0 ?

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We have to solve for x : (lg x^3)^2  - 6* lg x + 1 = 0

(lg x^3)^2  - 6* lg x + 1 = 0

=> (3 lg x)^2 - 6 lg x + 1 = 0

=> 9 (lg x)^2 - 6 lg x + 1 = 0

let lg x = y

=> 9y^2 - 6y + 1 = 0

=> ( 3y - 1)^2 = 0

=> 3y - 1 = 0

=> y = 1/3

lg x = 1/3

x = 10^(1/3)

The required result is x = 10^(1/3)

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lg^2 x^3 - 6lg x + 1 = 0

We will rewrite lg^2 x^3 = [ lg x^3 ]^2 = [ 3lg x]^2 = 9*lg^2 x

Now we will substitute into the equation.

==> 9*lg^2 x - 6lg x + 1 = 0

Now we will assume that lg x = y

==> 9y^2 - 6y + 1 = 0

Now let us factor the quadratic equation.

==> (3y-1)(3y-1) = 0

==> y= 1/3 ==> lg x = 1/3

We will rewrite into the exponent form.

==> x = 10^1/3

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