Question 328989
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x)\ =\ 2\ +\ \ln(x\ +\ 1)]


Ah ha!  You wrote the problem incorrectly the first time, didn't you?


Put the ln functions on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x)\ -\ \ln(x\ +\ 1)\ =\ 2]


The difference of the logs is the log of the quotient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{x}{x\ +\ 1}\right)\ =\ 2]


Use the definition of the logarithm function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{x\ +\ 1}\ =\ e^2]


Now just solve for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ e^2(x\ +\ 1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ e^2x\ +\ e^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ e^2x\ =\ e^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(1\ -\ e^2)\ =\ e^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{e^2}{1\ -\ e^2}]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^2\ >\ 1]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{e^2}{1\ -\ e^2}\ < 0]


and the solution must be excluded because of the domain restrictions of the log function.  Solution set is the empty set.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>