# What is x in (lgx)^2+ lg(x^5)-6=0?

You need to rewrite the term lg(x^5) using the power property of logarithms, such that:

lg^2 x + 5lg x - 6 = 0

You should come up with the notation  lg x = y , such that:

y^2 + 5y - 6 = 0

You need to use quadratic formula, such that:

y_(1,2) = (-5+-sqrt(25 + 24))/2 => y_(1,2) = (-5+-sqrt49)/2

y_1 = 1; y_2 = -6

You need to solve for x the following equations, such that:

lg x = 1 => x = 10^1 => x = 10

lg x = -6 => x = 10^(-6) => x = 1/10^6

Hence, evaluating the solutions to the given equation, yields x = 1/10^6 , x = 10.

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