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

Question 251163: How would I solve the a natural log problem like the one below?
Thank you!!!
((lnX)^2)-6 = -5lnX
Answer by Theo(13342) (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

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