Question 957761: Either factor or use the quadratic formula to solve the given equation. (Enter your answers as a comma-separated list.)
(ln x)^2 + 2 ln x = 3
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sometimes it helps to make your expression equal to a variable so the process of factoring doesn't look so messy.
if you let y = ln(x), then your equation becomes:
y^2 + 2y = 3
subtract 3 from both sides to get:
y^2 + 2y - 3 = 0
the equation is now in standard form of ay^2 + by + c = 0, where:
a = 1
b = 2
c = -3
this allows you to use the quadratic formula if you need it.
in this case you don't need it because the equation can be factored manually.
this equation can be factored as (y+3) * (y-1) = 0
set each of these factors to 0 and you get (y+3) = 0 and (y-1) = 0
solve for y to get y = -3 and y = 1
replace y with ln(x) that you made it equal to in the beginning.
y = -3 becomes ln(x) = -3
y = 1 becomes ln(x) = 1
solve for x in each of these equations.
ln(x) = -3 if and only if e^(-3) = x which results in x = .0497870684.
ln(x) = 1 if and only if e^1 = x which results in x = 2.718281828.
replace each of these values for x in the original equation to see if those original equations hold true.
the original equation is ln(x)^2 + 2*ln(x) = 3
replace x with .0497870684 and you get ln(x)^2 + 2*ln(x) = 3 becomes 3 = 3.
replace x with 2.718281828 and you get ln(x)^2 + 2*ln(x) = 3 becomes 3 = 3.
both values of x are good so they are your solutions.
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