Question

Need help: ln x + ln (x-6) = 2

Answer 1

Can you please explain how you do it etc, because I would rather understand it than just get a straight answer

Answer 2

I haven't taken a math class in a few months, so I forgot some of the material

Answer 3

We have:
\[
\ln x+\ln(x-6)=2\implies\\
\ln x(x-6)=2\implies\\
e^{\ln x(x-6)}=e^2
\](Considering the sums of logs and their definition):
\[
x(x-6)=e^2\implies\\
x^2-6x=e^2\implies\\
x^2-6x-e^2=\;0
\]Using quadratic formula we get:
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{36-4e^2}}{2}=3\pm\sqrt{9+e^2}
\]The solution cannot be negative, since \(\ln x\) is not defined as real for negative \(x\), so:
\[
x=3+\sqrt{9+e^2}
\]Et voilá.

Answer 4

Ah okay. I got to the x^(2) - 6x - e^(2) = 0 part but I didnt know what to do from there. Thank you so much :)

Answer 5

Sure thing.