Question 328988
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x)\ =\ \ln(2)\ +\ \ln(x\ +\ 1)]


Use:  The sum of the logs is the log of the product,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


to write:


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


Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(\alpha)\ =\ \ln(\beta)\ \ \Leftrightarrow\ \ \alpha\ =\ \beta]


to write:


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


Then solve for *[tex \Large x]


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


However, this solution must be discarded because the domain of the logarithm function is *[tex \LARGE \left(0,\infty\right)]


The solution set of this equation is the empty set.


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