Question 575448
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ \log\left(3x\right)\ =\ \log\left(\frac{x}{12}\right)\ +\ 2]


Add *[tex \LARGE -2] to both sides and add *[tex \LARGE \log\left(3x\right)] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{x}{12}\right)\ +\ \log\left(3x\right)\ =\ 0]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log\left(\frac{3x^2}{12}\right)\ =\ 0]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(1)\ =\ 0]


and


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


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x^2}{12}\ =\ 1]


Solve the quadratic, but discard the negative root because the domain of the log function is the positive Reals.


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