Thread: Exponential and Logarithmic Equations

Solve the logarithmic equation algebraically and approximate the result

Ln x + Ln(x - 2) = 1

I solved it to approximately 2.928

I want to ask why the negative number is extraneous because I'm having a difficult time grasping what is going on with the 2 variables being found on the graph, and graphing it wise. Seriously...why are negative answers extraneous???

Thanks in advance!

Originally Posted by ugkwan

Solve the logarithmic equation algebraically and approximate the result

Ln x + Ln(x - 2) = 1

I solved it to approximately 2.928

I want to ask why the negative number is extraneous because I'm having a difficult time grasping what is going on with the 2 variables being found on the graph, and graphing it wise. Seriously...why are negative answers extraneous???

Thanks in advance!

The left hand side can be written as Ln(x(x - 2)) using the usual log rule. Therefore you have the quadratic equation x^2 - 2x - e = 0. The gaps are left for you to fill, as is the exact solution. Note that only values of x that satisfy x > 2 will stisfy the original equation because the log of a negative number is not real.