Question 910912
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(x\ -\ 2)\ =\ 3\ -\ \log_3(x\ +\ 4)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(x\ -\ 2)\ +\ \log_3(x\ +\ 4)\ =\ 3]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3\left((x\ -\ 2)(x\ +\ 4)\right)\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_3(x^2\ +\ 2x\ -\ 8)\ =\ 3]


Since *[tex \Large \log_b(x)\ =\ y] if and only if *[tex \Large b^y\ =\ x],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ 8\ =\ 3^3\ =\ 27]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ 35 =\ 0]


Solve the quadratic for *[tex \Large x].  Discard the negative root because it is not in the domain of *[tex \Large \log_3(x\ -\ 2)] nor is it in the domain of *[tex \Large \log_3(x\ +\ 4)]


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