# SOLUTION: How would I solve the a natural log problem like the one below? Thank you!!! ((lnX)^2)-6 = -5lnX

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: How would I solve the a natural log problem like the one below? Thank you!!! ((lnX)^2)-6 = -5lnX      Log On

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 Question 251163: How would I solve the a natural log problem like the one below? Thank you!!! ((lnX)^2)-6 = -5lnXAnswer by Theo(3458)   (Show Source): You can put this solution on YOUR website!problem is: (ln(x))^2 - 6 = -5*ln(x) add 5*ln(x) to both sides of this equation to get: (ln(x))^2 + 5*ln(x) - 6 = 0 let y = ln(x) to get: y^2 + 5y - 6 = 0 factor this equation to get: (y+6)*(y-1) = 0 solve for y to get: y = -6 or y = 1 substitute ln(x) for y to get ln(x) = -6 or ln(x) = 1 by the laws of logarithms, ln(x) = y if and only if e^y = x based on that law, we get: ln(x) = -6 if and only if e^-6 = x and we get: ln(x) = 1 if and only if e^1 = x we will solve for each separately. first we'll solve for e^-6 = x using your calculator, e^-6 = .002478752 which makes: x = .002478752. next we'll solve for e^1 = x using your calculator, e^1 = 2.718281828 which makes: x = 2.718281828 x = .002478752 or x = 2.718281828 we can confirm by substituting in the original equations. it is: (ln(x))^2 - 6 = -5*ln(x) for x = 2.718281828, this equation becomes: (ln(2.718281828))^2 - 6 = -5*ln(2.718281828) since ln(2.718281828) = 1, this equation becomes: 1^2 - 6 = -5 which is true, confirming that x = 2.718281828 is good. for x = .002478752, this equation becomes: (ln(.002478752))^2 - 6 = -5*ln(.002478752) since ln(.002478752 = -6, this equation becomes: (-6)^2 - 6 = -5*-6 which becomes: 36 - 6 = 30 which is true, confirming that x = .002478752 is good. both answers are good. your answer is: x = 2.718281828 or: x = .002478752