SOLUTION: Ln(x)+Ln(x-2)=1

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Question 39229This question is from textbook Algebra 2
: Ln(x)+Ln(x-2)=1 This question is from textbook Algebra 2

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
Ln(x)+Ln(x-2)=1
Ln((x)(x-2))=1
Ln(x^2-2x)=1
x^2-2x=e
x^2-2x-e=0
Looks a little like what is below. No solutions exist.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B2.718281828+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A2.718281828=-6.873127312.

The discriminant -6.873127312 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -6.873127312 is + or - sqrt%28+6.873127312%29+=+2.62166498851398.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B2.718281828+%29