Need help with Logs

[captaininsano]

I have been trying to solve this one but have not been able to. Perhaps I have been thinking about the problem wrong. Can you help? Here it is:

Ln(2x+3)-Ln(x-1)=6

Thanks in advance.

 

[pka]

I have been trying to solve this one but have not been able to. Perhaps I have been thinking about the problem wrong. Can you help? Here it is:
Ln(2x+3)-Ln(x-1)=6

That is equivalent to \(\displaystyle \dfrac{2x+3}{x-1}=e^6.\)

 

[captaininsano]

That is equivalent to \(\displaystyle \dfrac{2x+3}{x-1}=e^6.\)

So from there I get 2x+3=(x-1)(e^6)

then: Ln(2x+3)=Ln(x-1)+6

then what? I have no idea.

 

[pka]

So from there I get 2x+3=(x-1)(e^6) YES

then: Ln(2x+3)=Ln(x-1)+6 NO!

\(\displaystyle (e^6-2)x=3+e^6\) now solve for \(\displaystyle x~.\)