Question 975203
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\cdot \ln(x\ +\ 1)\ =\ \ln(x^2\ -\ 1)\ +\ \ln(5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x\ +\ 1)^2\ -\ \ln\left((x\ -\ 1)(x\ +\ 1)\right)\ =\ \ln(5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{(x\ +\ 1)^2}{(x\ -\ 1)(x\ +\ 1)}\right)\ =\ \ln(5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(\frac{x\ +\ 1}{x\ -\ 1}\right)\ =\ \ln(5)]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(a)\ =\ \ln(b)\ \Leftrightarrow\ a\ =\ b\ \forall a,\, b\ \in\ \text{dom}\left{\ln(x)\right}]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ +\ 1}{x\ -\ 1}\ =\ 5]


You can take it from here, I should imagine.


Don't forget to make sure that *[tex \Large x\ +\ 1] and *[tex \Large x^2\ -\ 1] are actually in the domain of the log function before you publish your solution.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \