Question 251163
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